# NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry (Ex 7.4) Exercise 7.4

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## Access NCERT Solutions for Class-10 Maths Chapter 7 – Coordinate Geometry

Exercise 7.4

1. Determine the ratio in which the line $\text{2x+y-4=0}$  divides the line segment joining the points $\text{A}\left( \text{2,-2} \right)$  and $\text{B}\left( \text{3,7} \right)$ .

Ans: Given  the points $\text{A}\left( \text{2,-2} \right)$ and $\text{B}\left( \text{3,7} \right)$

Given  the line $\text{2x+y-4=0}$.

• Let the given line $\text{2x+y-4=0}$ divide the line segment joining the points $\text{A}\left( \text{2,-2} \right)$  and $\text{B}\left( \text{3,7} \right)$  in a ratio $\text{k:1}$  at the point $\text{C}$ .

(Image will be uploaded soon)

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$.

Therefore following the formula Coordinates of the point of division ,$\text{C}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{3k+2}}{\text{k+1}}\text{,}\dfrac{\text{7k-2}}{\text{k+1}} \right)$

This point of division $\text{C}$also lies on the line$\text{2x+y-4=0}$……(1)

By substituting the coordinates value of $C\left( x,y \right)$ in equation(1), we get,

$\text{2}\left( \dfrac{\text{3k+2}}{\text{k+1}} \right)\text{+}\left( \dfrac{\text{7k-2}}{\text{k+1}} \right)\text{-4=0}$

$\Rightarrow \dfrac{6k+4+7k-2-4k-4}{k+1}=0$

By cross multiplication and further computing,

$\Rightarrow 9k-2=0$

$\Rightarrow k=\dfrac{2}{9}$

Therefore, the line $\text{2x+y-4=0}$ divides the line segment joining the points $\text{A}\left( \text{2,-2} \right)$  and $\text{B}\left( \text{3,7} \right)$ in the ratio$2:9$ internally.

2. Find a relation between $\text{x}$  and $\text{y}$ if the points $\left( \text{x,y} \right)$ ,$\left( \text{1,2} \right)$ and $\left( \text{7,0} \right)$ are collinear.

Ans: Given the points$\left( \text{x,y} \right)$,$\left( \text{1,2} \right)$and$\left( \text{7,0} \right)$. Let us denote the given points by $\text{A,B,C}$ respectively. Then ,

• Let $\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{x,y} \right)$

• Let $\text{B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{1,2} \right)$

• Let  $\text{C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{7,0} \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$

For collinear points, area of the triangle formed by those points is $\text{0}$ .

Therefore, we have,

for collinear points,   $\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}\text{=0}$   …….(1)

By substituting the values of vertices from the given points$\text{A,B,C}$ in the Equation (1), we get,

Area of $\vartriangle \text{ABC}$$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{x}\left( \text{2-0} \right)\text{+1}\left( \text{0-y} \right)\text{+7}\left( \text{y-2} \right) \right] \Rightarrow \text{0=}\dfrac{\text{1}}{\text{2}}\left[ \text{2x-y+7y-14} \right] Simplifying, we get, \dfrac{\text{1}}{\text{2}}\left[ \text{2x+6y-14} \right]\text{=0} \Rightarrow \text{2x+6y-14=0} \Rightarrow \text{x+3y-7=0} Then the required relation between x and y is given by \text{x+3y-7=0} . 3. Find the centre of a circle passing through the points \left( \text{6,-6} \right) ,\left( \text{3,-7} \right) and \left( \text{3,3} \right) . Ans: Given the points \left( \text{6,-6} \right)\text{,}\left( \text{3,-7} \right)and \left( \text{3,3} \right) passing through a circle. Let us denote the given points by \text{A,B,C} respectively. Then , • Let P\left( \text{x,y} \right)be the centre of the circle passing through the points \left( \text{6,-6} \right)\text{,}\left( \text{3,-7} \right)and \left( \text{3,3} \right). • The points \text{A}\left( \text{6,-6} \right)\text{,B}\left( \text{3,-7} \right)and \text{C}\left( \text{3,3} \right) are the points on the circumference of the circle. (Image will be uploaded soon) Now , consider the distance formula, Distance between a pair of points is given by \sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}} . Distance from centreP\left( \text{x,y} \right) toA ,B and C are as following, using the Distance formula, \text{PA=}\sqrt{{{\left( \text{x-6} \right)}^{\text{2}}}\text{+}{{\left( \text{y+6} \right)}^{\text{2}}}} \text{PB=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y+7} \right)}^{\text{2}}}} \text{PC=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-3} \right)}^{\text{2}}}} From the figure that \text{PA=PB}, since \text{PA} and \text{PB} are the radii of the same circle. \sqrt{{{\left( \text{x-6} \right)}^{\text{2}}}\text{+}{{\left( \text{y+6} \right)}^{\text{2}}}}\text{=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y+7} \right)}^{\text{2}}}} {{\text{x}}^{\text{2}}}\text{+36-12x+}{{\text{y}}^{\text{2}}}\text{+36+12y=}{{\text{x}}^{\text{2}}}\text{+9-6x+}{{\text{y}}^{\text{2}}}\text{+49+14y} Simplifying, \text{-6x-2y+14=0} \Rightarrow 3x+y=7…… (1) Similarly PA=PC since \text{PA} and \text{PC} are the radii of same circle. \sqrt{{{\left( \text{x-6} \right)}^{\text{2}}}\text{+}{{\left( \text{y+6} \right)}^{\text{2}}}}\text{=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-3} \right)}^{\text{2}}}} {{\text{x}}^{\text{2}}}\text{+36-12x+}{{\text{y}}^{\text{2}}}\text{+36+12y=}{{\text{x}}^{\text{2}}}\text{+9-6x+}{{\text{y}}^{\text{2}}}\text{+9-6y} \Rightarrow \text{-6x+18y+54=0} \Rightarrow \text{-3x+9y=-27} ……(2) On adding equation (1) and (2), we get, \text{10y=-20} \text{y=-2} Substituting the value of \text{y} in equation (1) , we obtain \text{3x-2=7} \text{3x=9} \text{x=3} Therefore , the coordinates of the centreP\left( \text{x,y} \right) of the circle is \left( \text{3,-2} \right) . 4. The two opposite vertices of a square are\left( \text{-1,2} \right) and \left( \text{3,2} \right). Find the coordinates of the other two vertices. Ans: Given two opposite vertices of a square\left( \text{-1,2} \right)and\left( \text{3,2} \right). Let us denote the given points by \text{A,C} respectively. Then , • Let ABCD be a square having vertices\text{A}\left( \text{-1,2} \right) and \text{C}\left( \text{3,2} \right). • Let B\left( x,y \right)and D\left( {{x}_{1}},{{y}_{1}} \right) be the unknown vertices. (Image will be uploaded soon) Since the sides of a square are equal to each other. We have, \text{AB=BC} Now , consider the distance formula, Distance between a pair of points is given by \sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}} . By the distance formula the distance \text{AB} is given by, \text{AB=}\sqrt{{{\left( \text{x+1} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}} And the distance \text{BC} is given by , \text{BC=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}} Therefore the condition \text{AB=BC} implies, \sqrt{{{\left( \text{x+1} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}}\text{=}\sqrt{{{\left( \text{x-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}} Squaring both sides and expanding, {{\text{x}}^{\text{2}}}\text{+2x+1+}{{\text{y}}^{\text{2}}}\text{-4y+4=}{{\text{x}}^{\text{2}}}\text{+9-6x+}{{\text{y}}^{\text{2}}}\text{+4-4y} \Rightarrow 8x=8 \Rightarrow x=1 Then coordinates of \text{B} is \left( \text{1,y} \right) . Now we know, all the interior angles are of {{90}^{\circ }} in a square. Then by Pythagoras theorem in the triangle \vartriangle ABC, A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} Using distance formula, the distance \text{AC} is given by \text{AC=}\left( \sqrt{{{\left( \text{3+1} \right)}^{\text{2}}}\text{+}{{\left( \text{2-2} \right)}^{\text{2}}}} \right) Therefore the condition A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} implies {{\left( \sqrt{{{\left( \text{1+1} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}} \right)}^{\text{2}}}\text{+}{{\left( \sqrt{{{\left( \text{1-3} \right)}^{\text{2}}}\text{+}{{\left( \text{y-2} \right)}^{\text{2}}}} \right)}^{\text{2}}}\text{=}{{\left( \sqrt{{{\left( \text{3+1} \right)}^{\text{2}}}\text{+}{{\left( \text{2-2} \right)}^{\text{2}}}} \right)}^{\text{2}}} Simplifying and expanding, \text{4+}{{\text{y}}^{\text{2}}}\text{+4-4y+4+}{{\text{y}}^{\text{2}}}\text{-4y+4=16} \Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+16-8y=16} \Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{-8y=0} \Rightarrow \text{y}\left( \text{y-4} \right)\text{=0} \Rightarrow \text{y=0 or 4} Hence the required unknown verticles of the square \text{ABCD} are B\left( 1,0 \right) and D\left( 1,4 \right). 5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of \text{1 m} from each other. There is a triangular grassy lawn in the plot as shown in the following figure. The students are to sow seeds of flowering plants on the remaining area of the plot. (Image will be uploaded soon) (i)Taking \text{A} as origin, find the coordinates of the vertices of the triangle. Also calculate the area of the triangle in this case. Ans: • If we take \text{A} as origin, the line segment \text{AD} will represent the \text{x-axis} and AB will represent the \text{y-axis}. • Here it is mentioned that saplings of gulmohar are planted \text{1 m} apart from each other along \text{AD} and AB. • Then it is obvious from figure that the coordinates of point P , Qand R are \left( 4,6 \right) ,\left( 3,2 \right) and \left( 6,5 \right) respectively. Then we have, P\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,6 \right) Q\left( {{x}_{2}},{{y}_{2}} \right)=\left( 3,2 \right) R\left( {{x}_{3}},{{y}_{3}} \right)=\left( 6,5 \right) Consider the formula for area of a triangle whose vertices are \left( {{x}_{1}},{{y}_{1}} \right) ,\left( {{x}_{2}},{{y}_{2}} \right) and \left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right) , we have, Area of a triangle\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\} …… (1) By substituting the values of vertices P , Qand Robtained in the Equation (1), Area of \vartriangle \text{PQR}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{4}\left( \text{2-5} \right)\text{+3}\left( \text{5-6} \right)\text{+6}\left( \text{6-2} \right) \right]$

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{-12-3+24} \right]$

$\text{=}\dfrac{\text{9}}{\text{2}}\text{ square units}$

(ii) What will be the coordinates of the vertices of $\vartriangle \text{PQR}$ if $\text{C}$  is the origin? Also calculate the areas of the triangle in this case. What do you observe?

Ans:

• If we take $C$ as origin, $\text{CB}$ will represent the $\text{x-axis}$ , and $\text{CD}$  will represent the $\text{y-axis}$ .

• Here it is mentioned that saplings of gulmohar are planted $\text{1 m}$ apart from each other along $\text{CD}$ and  $\text{CB}$.

• Then it is obvious from the figure 5 that the coordinates of vertices $P$ , $Q$and $R$ are $\left( 12,2 \right)$ ,$\left( 13,6 \right)$and $\left( 10,3 \right)$ respectively. Then we have,$\text{P}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{12,2} \right)$

$\text{Q}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{13,6} \right)$

$\text{R}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{10,3} \right)$

Consider the formula for area of a triangle whose vertices are $\left( {{x}_{1}},{{y}_{1}} \right)$ ,$\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)$ , we have,

Area of a triangle$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\}$ …… (1)

By substituting the values of vertices$P$ , $Q$and $R$  obtained in the equation (1),

Area of triangle $\vartriangle \text{PQR}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\} \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{12}\left( \text{6-3} \right)\text{+13}\left( \text{3-2} \right)\text{+10}\left( \text{2-6} \right) \right] \text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{36+13-40} \right] \text{=}\dfrac{\text{9}}{\text{2}}\text{ square units} It can be observed that the area of the triangle is same in both the cases i.e taking \text{A} as origin and taking \text{C} as origin. 6. The vertices of a \vartriangle \text{ABC} are \text{A}\left( \text{4,6} \right),\text{B}\left( \text{1,5} \right) and \text{C}\left( \text{7,2} \right) . A line is drawn to intersect sides \text{AB} and \text{AC} at \text{D} and \text{E} respectively, such that \dfrac{\text{AD}}{\text{AB}}=\dfrac{\text{AE}}{\text{AC}}=\dfrac{\text{1}}{\text{4}} Calculate the area of the \vartriangle \text{ADE} and compare it with the area of the triangle \text{ABC} . (Recall Converse of basic proportionality theorem and Theorem 6.6 related to Ratio of areas of two similar triangles) Ans: Given the points \text{A}\left( \text{4,6} \right),\text{B}\left( \text{1,5} \right)and\text{C}\left( \text{7,2} \right). Given that \dfrac{\text{AD}}{\text{AB}}\text{=}\dfrac{\text{AE}}{\text{AC}}\text{=}\dfrac{\text{1}}{\text{4}} (Image will be uploaded soon) Therefore the points \text{D} and \text{E} divide \text{AB} and \text{AC} respectively in the ratio 1:3 . By section formula, we have, \text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right) , where \text{P}\left( \text{x,y} \right) divides the join of\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right) and \left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right) in the ratio \text{m:n}. Therefore Coordinates of the point D=\left( \dfrac{1\times 1+3\times 4}{1+3},\dfrac{1\times 5+3\times 6}{1+3} \right) =\left( \dfrac{13}{4},\dfrac{23}{4} \right) And Coordinates of the point E=\left( \dfrac{1\times 7+3\times 4}{1+3},\dfrac{1\times 2+3\times 6}{1+3} \right) =\left( \dfrac{19}{4},\dfrac{20}{4} \right) Consider the formula for area of a triangle whose vertices are \left( {{x}_{1}},{{y}_{1}} \right) ,\left( {{x}_{2}},{{y}_{2}} \right) and \left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right) , we have, Area of a triangle\text{=}\dfrac{\text{1}}{\text{2}}\left\{ {{\text{x}}_{\text{1}}}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{3}}} \right)\text{+}{{\text{x}}_{\text{2}}}\left( {{\text{y}}_{\text{3}}}\text{-}{{\text{y}}_{\text{1}}} \right)\text{+}{{\text{x}}_{\text{3}}}\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right) \right\} …… (1) By substituting the vertices \text{A,D,E} obtained in equation (1), Area of \vartriangle \text{ADE}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ 4\left( \dfrac{23}{4}\text{-}\dfrac{20}{4} \right)\text{+}\dfrac{13}{4}\left( \dfrac{20}{4}\text{-6} \right)\text{+}\dfrac{19}{4}\left( \text{6-}\dfrac{23}{4} \right) \right\}$

$\text{=}\dfrac{\text{1}}{\text{2}}\left[ \text{3-}\dfrac{\text{13}}{\text{4}}\text{+}\dfrac{\text{19}}{\text{16}} \right]$

$\text{=}\dfrac{\text{15}}{\text{32}}\text{ }square\text{ unit}$

Substituting the vertices $\text{A,B,C}$ from the given points in equation (1),

Area of $\vartriangle \text{ABC}$$\text{=}\dfrac{\text{1}}{\text{2}}\left\{ 4\left( \text{5-2} \right)\text{+1}\left( \text{2-6} \right)\text{+7}\left( \text{6-5} \right) \right\} =\dfrac{15}{2} Therefore , \dfrac{\text{area of }\vartriangle \text{ADE}}{area\text{ of }\vartriangle \text{ABC}}=\dfrac{\dfrac{15}{32}}{\dfrac{15}{2}} =\dfrac{1}{16} Therefore the area of \vartriangle \text{ABC} is 16 times the area of \vartriangle \text{ADE}. Alternatively, if two sides of a triangle is divided by a line segment in the same ratio then the line segment is parallel to the third side of the triangle and the two triangles formed bt this line segment is similar to the original. Therefore the ratio between the areas of these two triangle will be the square of the ratio of the sides of these two triangles formed. Hence the ratio of the area of \vartriangle \text{ADE} to the area of \vartriangle \text{ABC} is {{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{1}{16} . 7. Let \text{A}\left( \text{4,2} \right),\text{B}\left( \text{6,5} \right)and \text{C}\left( \text{1,4} \right) be the vertices of \vartriangle \text{ABC} (i)The median from \text{A} meets \text{BC} at \text{D } . Find the coordinates of point \text{D }. (Image will be uploaded soon) Ans: Given the points\text{A}\left( \text{4,2} \right) ,\text{B}\left( \text{6,5} \right) and\text{C}\left( \text{1,4} \right)are the vertices of a triangle. \text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{=}\left( \text{4,2} \right) \text{B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\text{=}\left( \text{6,5} \right) \text{C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\text{=}\left( \text{1,4} \right) Here it is given that the median from \text{A} meets \text{BC} at \text{D }. Then \text{AD} is median of the triangle~ABC. Hence D is the midpoint of \text{BC} . By section formula, we have, \text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right) , where \text{P}\left( \text{x,y} \right) divides the join of\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right) and \left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right) in the ratio \text{m:n}. Now Ddivides the line segment \text{BC} in the ratio \text{1:1} Therefore Coordinates of D$$\text{=}\left( \dfrac{\text{6+1}}{\text{2}}\text{,}\dfrac{\text{5+4}}{\text{2}} \right)\text{=}\left( \dfrac{\text{7}}{\text{2}}\text{,}\dfrac{\text{9}}{\text{2}} \right)$

(ii) Find the coordinates of the point $\text{P}$  on $\text{AD}$  such that $\text{AP:PD=2:1}$

Ans: Given that $\text{AP:PD=2:1}$

Hence the Point $\text{P}$  divides the side $AD$ in a ratio$2:1$.

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$.

Therefore Coordinates of $\text{P}$is given by,

$\text{P=}\left( \dfrac{\text{2 }\!\!\times\!\!\text{ }\dfrac{\text{7}}{\text{2}}\text{+1 }\!\!\times\!\!\text{ 4}}{\text{2+1}}\text{,}\dfrac{\text{2 }\!\!\times\!\!\text{ }\dfrac{\text{9}}{\text{2}}\text{+1 }\!\!\times\!\!\text{ 2}}{\text{2+1}} \right)\text{=}\left( \dfrac{\text{11}}{\text{3}}\text{,}\dfrac{\text{11}}{\text{3}} \right)$

(iii) Find the coordinates of point $\text{Q}$  and $\text{R}$  on medians $\text{BE}$  and $\text{CF}$  respectively such that $\text{BQ: QE = 2:1}$ and $\text{CR: RF = 2:1}\text{.}$

Ans: Since  $\text{BE}$is the median of the triangle$~ABC$,$E$ is the mid-point of side $AC$ . Then $\text{BE}$ divides the side $AC$in the ratio $\text{1:1}$.

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$.

Hence ,Coordinates of $E$ is given by,

$E=\left( \dfrac{4+1}{2},\dfrac{2+4}{2} \right)=\left( \dfrac{5}{2},3 \right)$

Now  given that $\text{BQ: QE = 2:1}$

Then thepoint $Q$ divides the side $BE$ in a ratio$2:1$ .

Hence the Coordinates $Q$of is given by,

$Q=\left( \dfrac{2\times \dfrac{5}{2}+1\times 6}{2+1},\dfrac{2\times 3+1\times 5}{2+1} \right)=\left( \dfrac{11}{3},\dfrac{11}{3} \right)$

Since $CF$is the median of the triangle $~ABC$ ,$\text{F}$  is the mid-point of side$AB$. Hence , $CF$will divide the side $AB$ in the ratio $\text{1:1}$.

Therefore Coordinates of $\text{F}$ is given by,

$F=\left( \dfrac{4+6}{2},\dfrac{2+5}{2} \right)=\left( 5,\dfrac{7}{2} \right)$

Now  given that $\text{CR: RF = 2:1}\text{.}$

Then the point $\text{R}$  divides the side $\text{CF}$  in a ratio $\text{2:1}$ .

Then coordinates of $\text{R}$is given by,

$R=\left( \dfrac{2\times 5+1\times 1}{2+1},\dfrac{2\times \dfrac{7}{2}+1\times 4}{2+1} \right)=\left( \dfrac{11}{3},\dfrac{11}{3} \right)$

(iv) What do you observe?

Ans: From the obtained values of $\text{P,Q,R}$it can be observed that, the coordinates of the points $\text{P,Q,R}$ are the same. Therefore, each of these points represent the same point on the plane i.e., the centroid of the triangle.

(v) If$\text{A(x1, y1)}$ , $\text{B}\left( \text{x2, y2} \right)$ and $\text{C}\left( \text{x3, y3} \right)$ are the vertices of$\vartriangle \text{ABC}$ , find the coordinates of the centroid of the triangle.

Ans: Given that$A\left( {{x}_{1}},\text{ }{{y}_{1}} \right)$ , $B\left( {{x}_{1}},{{y}_{2}} \right)$ and $C\left( {{x}_{3}},{{y}_{3}} \right)$ are the vertices of the triangle $\text{ABC}$ .The median $AD$ of the triangle divides the side $BC$ in the ratio $\text{1:1}$. Therefore, $D$  is the mid-point of side $BC$ .

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$.

Then Coordinates of$D$ is given by,

$D=\left( \dfrac{{{x}_{2}}+{{x}_{3}}}{2},\dfrac{{{y}_{2}}+{{y}_{3}}}{2} \right)$

Let the centroid of the triangle $\text{ABC}$be $O$ .

Then the  Point $O$  divides the side $AD$ in the ratio $2:1$.

Then by the section formula, coordinates of  $O$is given by,

$O=\left( \dfrac{2\times \dfrac{{{x}_{2}}+{{x}_{3}}}{2}+1\times {{x}_{1}}}{2+1},\dfrac{2\times \dfrac{{{y}_{2}}+{{y}_{3}}}{2}+1\times {{y}_{1}}}{2+1} \right)$

$=\left(\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)$.

8. $\text{ABCD}$ is a rectangle formed by the points$\text{A}\left( \text{-1,-1} \right)$ , $\text{B}\left( \text{-1,4} \right)$ ,$\text{ }\!\!~\!\!\text{ C}\left( \text{5,4} \right)$  and $\text{D}\left( \text{5,-1} \right)$ . $\text{P}$ , $\text{Q}$ , $\text{R}$  and $\text{S}$  are the mid-points of $\text{AB}$ , $\text{BC}$ ,$\text{ }\!\!~\!\!\text{ CD}$  , and $\text{DA}$ respectively. Is the quadrilateral $\text{PQRS}$ is a square? A rectangle? Or a rhombus? Justify your answer.

Ans: Given the points$\text{A}\left( \text{-1,-1} \right)$ ,$\text{B}\left( \text{-1,4} \right)$ ,$\text{C}\left( \text{5,4} \right)$,and $D\left( \text{5,-1} \right)$ are the vertices of a rectangle.

Given $P$ is the midpoint of $AB$

By section formula, we have,

$\text{P}\left( \text{x,y} \right)\text{=}\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{+n}{{\text{x}}_{\text{1}}}}{\text{m+n}}\text{,}\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{+n}{{\text{y}}_{\text{1}}}}{\text{m+n}} \right)$  , where  $\text{P}\left( \text{x,y} \right)$ divides the join of$\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{x}}_{\text{2}}} \right)$ and $\left( {{\text{y}}_{\text{1}}}\text{,}{{\text{y}}_{\text{2}}} \right)$  in the ratio $\text{m:n}$.

Now $P$divides  the line segment $AB$ in the ratio $\text{1:1}$.

Therefore , the coordinates  of $P$ are $\left( \dfrac{\text{-1-1}}{\text{2}}\text{,}\dfrac{\text{-1+4}}{\text{2}} \right)\text{=}\left( \text{-1,}\dfrac{\text{3}}{\text{2}} \right)$

Similarly, the coordinates of $\text{Q}$,$R$ and $S$ are $\left( 2,4 \right)$ ,$\left( \text{5,}\dfrac{\text{3}}{\text{2}} \right)$ and $\left( 2,-1 \right)$respectively.

Now , consider the distance formula,

Distance between a pair of points is given by $\sqrt{{{\left( {{\text{x}}_{\text{1}}}\text{-}{{\text{x}}_{\text{2}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{y}}_{\text{1}}}\text{-}{{\text{y}}_{\text{2}}} \right)}^{\text{2}}}}$ where $\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)$ and $\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)$ are the coordinates of the given pair of points.

Distance between the pair of  points $P$ and $Q$ is given by,

$\text{PQ=}\sqrt{{{\left( \text{-1-2} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{3}}{\text{2}}\text{-4} \right)}^{\text{2}}}}$

$=\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$=\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of  points $Q$ and $R$ is given by,

$\text{QR=}\sqrt{{{\left( \text{2-5} \right)}^{\text{2}}}\text{+}{{\left( \text{4-}\dfrac{\text{3}}{\text{2}} \right)}^{\text{2}}}}$

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of  points $R$ and $S$ is given by,

$\text{RS=}\sqrt{{{\left( \text{5-2} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{3}}{\text{2}}\text{+1} \right)}^{\text{2}}}}$

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of points$S$ and $P$ is given by,

$\text{SP=}\sqrt{{{\left( \text{2+1} \right)}^{\text{2}}}\text{+}{{\left( \text{-1-}\dfrac{\text{3}}{\text{2}} \right)}^{\text{2}}}}$

$\text{=}\sqrt{\text{9+}\dfrac{\text{25}}{\text{4}}}$

$\text{=}\sqrt{\dfrac{\text{61}}{\text{4}}}$

Distance between the pair of points $P$ and $R$ is given by,

$\text{PR=}\sqrt{{{\left( \text{-1-5} \right)}^{\text{2}}}\text{+}{{\left( \dfrac{\text{3}}{\text{2}}\text{-}\dfrac{\text{3}}{\text{2}} \right)}^{\text{2}}}}$

$\text{=6}$

Distance between the pair of points $Q$ and $S$ is given by,

$\text{QS=}\sqrt{{{\left( \text{2-2} \right)}^{\text{2}}}\text{+}{{\left( \text{4+1} \right)}^{\text{2}}}}$

$\text{=5}$

(Image will be uploaded soon)

It can be observed that all sides $\text{PQ,SP,QR,RS}$  of the given quadrilateral $\text{PQRS}$  are of the same length but the diagonals$\text{PR,QS}$ are of different lengths. Therefore, $\text{PQRS}$ is a rhombus.

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Exercise 7.4

Opting for the NCERT solutions for Ex 7.4 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.4 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 7 Exercise 7.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.4, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 10 Maths Chapter 7 Exercise 7.4 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs (Frequently Asked Questions)

1. How many questions are asked in the exercise of Class 10 Maths Chapter 7?

Exercise 7.4 of Class 10 Maths Chapter 7 contains 8 Questions which includes 6 long answer questions, 1 practical based question, 1 reasoning question. Solutions to these problems are also provided in the NCERT Solutions provided by Vedantu. These are carefully drafted to assist the student in scoring good marks in the examination. It provides you with much-needed problem-solving practice.

2. What are the topics covered under this Chapter 7 Exercise 7.4 of Class 10 Maths?

This chapter Coordinate Geometry deals with finding the area between two points whose coordinate values are provided and the topic covered under Chapter 7 Exercise 7.4 of Class 10 Maths is Section formula. Some of the solved problems are also provided in this chapter also.

3. Why are NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.4 considered as the best resource?

Opting for the NCERT solutions for Ex 7.4 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 10 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 10 Maths Chapter 7 Exercise 7.4 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

4. Can I download the NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.4 from Vedantu site?

Yes, you can. We have provided the Exercise 7.4 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

5. Why is Chapter 7 Exercise 7.4 of Class 10 Maths important?

Chapter 7 Exercise 7.4 is important because the sums given in Exercise 7.4 of Class 10 Maths are very tricky and demand the students to use their higher-order thinking and mathematical skills. This particular exercise is an interesting way to understand whether a student has an in-depth or only superficial understanding of a lesson. If you are able to solve NCERT Solutions for Class 10 Maths Exercise 7.4 of Chapter 7, you enhance your logical skills.

6. How much time do students need to complete Exercise 7.4 of Chapter 7 of Class 10 Mathematics?

It is important to follow a schedule with time properly divided between chapters to ace your Class 10 Maths board exams. This means you must know how much time you will require to finish each chapter. To finish Chapter 7, Exercise 7.4 of Class 10 Maths, students will require no more than 1 or 2 days. It will be easier for you to understand and complete the exercise if you have already gone through the NCERT solutions for Class 10 Maths free of cost by Vedantu

7. Is Exercise 7.4 of Chapter 7 of Class 10 Maths easy or difficult?

Exercise 7.4 of Chapter 7 in Class 10 Mathematics is neither very easy nor very difficult. This is because a few questions from the exercise may seem easy while a few of them might be difficult for you. To help you to understand the exercise well, and solve all the questions easily and without doubt, it's a good idea to download and take the help of the NCERT Solutions by Vedantu for Class 10 Maths, which have detailed answers to all the questions and exercises.

8. Do questions from Exercise 7.4 of Class 10 Mathematics come in Board exams?

Yes, questions are usually asked from Chapter 7, Exercise 7.4 In Class 10 Maths in the final Board Examination. Note here that the first question of Exercise 7.4 has been asked multiple times in the board exams and also has a good chance of being asked in the school level exams. This exercise includes several interesting and logical problems that require you to have a sound understanding of mathematical concepts and their applications. So, make sure that you practice this exercise thoroughly.

9. How are NCERT Solutions for Chapter 7 Exercise 7.4 of Class 10 Maths helpful for studying Mathematics?

NCERT Solutions for Chapter 7, Exercise 7.4 of Class 10 Maths are extremely helpful for studying Mathematics and have the following benefits:

• Consists of solutions for the long answer questions included in the given Exercise 7.4.

• Solving these NCERT sums and exercises can provide you with solid practice for good preparation for the exams.

• You will gain proper knowledge of the methodology of solving the problems related to the area of a triangle.

So, it's a good idea to download the NCERT Solutions available on the Vedantu app and website.

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