Find the point on the \[3x - 4y - 1{\text{ }} = 0\], where are at a distance of 5 units from the point (3, 2)
Answer
282.9k+ views
Hint: A straight line is a single line with no bending. And it extends from the minus infinite to plus infinite has no curvature is called a straight line and the standard form of the line is given below
\[y = mx + p\]
Where the m is known as the slope and the p is intercepted at the y axis which is matched with the given line equation. In this question, we have to find the point on the line having the distance of the five-unit from the given point.
Complete step-by-step solution:
Step 1:
Consider
First we write the given equation and point and distance D
\[3x - 4y - 1{\text{ }} = 0\]
\[D = 5\] From the given point to the point on the line \[\left( {u,v} \right)\] assume now D is given by the below distance formula so
\[D = \sqrt {{{\left( {u - {x_1}} \right)}^2} + {{\left( {v - {y_1}} \right)}^2}} \\
\text{We have} \\
\Rightarrow \left( {{x_1},{y_1}} \right) = (3,2)\,\,and\,\,D = 5\,\,put\,\,the\,\,value \\
\Rightarrow D = \sqrt {{{\left( {u - {x_1}} \right)}^2} + {{\left( {v - {y_1}} \right)}^2}} \\
\Rightarrow {D^2} = {\left( {u - {x_1}} \right)^2} + {\left( {v - {y_1}} \right)^2} \\
\Rightarrow {5^2} = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\text{now from the equation} 3x - 4y = 1 \text{satisfy the point (u,v) then} \\
\Rightarrow 3u - 4v = 1\, \\
\Rightarrow 3u = 1 + 4v\, \\
\Rightarrow u = \dfrac{{1 + 4v}}{3} \]
\[\text{Now put the value into} \\
\Rightarrow 25 = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{1 + 4v}}{3}\, - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{1 + 4v - 9}}{3}} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{4v - 8}}{3}} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = \left( {\dfrac{{16{v^2} + 64 - 64v}}{9}} \right) + \left( {{v^2} + 4 - 4v} \right) \\
\Rightarrow 25 \times 9 = \left( {16{v^2} + 64 - 64v} \right) + \left( {9{v^2} + 36 - 36v} \right) \\
\Rightarrow 225 = \left( {25{v^2} + 100 - 100v} \right) \\
\Rightarrow 25{v^2} + 100 - 100v = 225 \\
\Rightarrow 25{v^2} + 100 - 225 - 100v = 0 \\
\Rightarrow 25{v^2} - 125 - 100v = 0 \\
\Rightarrow 25\left( {{v^2} - 5 - 4v} \right) = 0 \\
\Rightarrow {v^2} - 4v - 5 = 0 \\
\Rightarrow {v^2} - 5v + v - 5 = 0 \\
\Rightarrow v\left( {v - 5} \right) + 1\left( {v - 5} \right) = 0 \\
\Rightarrow \left( {v + 1} \right)\left( {v - 5} \right) = 0 \\
or \\
\Rightarrow v = - 1\,\,or\,\,v = 5 \\
\text{now put it into the} \\
\Rightarrow u = \dfrac{{1 + 4v}}{3}\, \\
\text{then},v = - 1\,\, \\
\Rightarrow u = \dfrac{{1 + 4\left( { - 1\,} \right)}}{3}\, = \dfrac{{1 - 4}}{3}\, = \dfrac{{ - 3}}{3} = - 1 \\
u = - 1 \\
\text{ similarly for}\,v = 5 \\
\Rightarrow u = \dfrac{{1 + 4v}}{3} \\
then,v = 5\,\, \\
\Rightarrow u = \dfrac{{1 + 4\left( {5\,} \right)}}{3}\, = \dfrac{{1 + 20}}{3}\, = \dfrac{{21}}{3} = 7 \\
\Rightarrow u = 7 \\ \]
Now we got the pair of the points $\left( {u,v} \right) = \left( {7,5} \right)$ or $\left( {u,v} \right) = \left( { - 1, - 1} \right) $
Which is on the given line and our desired answer or point (7, 5), or (-1,-1)
Note: The straight line question is very straight forward we have just followed the instruction given by the question and we got the points on the line which are (7, 5), or (-1,-1), and both points on line having the distance from the point (3, 2) are 5 unite and can be cross-checked with the help of the distance formula
\[y = mx + p\]
Where the m is known as the slope and the p is intercepted at the y axis which is matched with the given line equation. In this question, we have to find the point on the line having the distance of the five-unit from the given point.
Complete step-by-step solution:
Step 1:
Consider
First we write the given equation and point and distance D
\[3x - 4y - 1{\text{ }} = 0\]
\[D = 5\] From the given point to the point on the line \[\left( {u,v} \right)\] assume now D is given by the below distance formula so
\[D = \sqrt {{{\left( {u - {x_1}} \right)}^2} + {{\left( {v - {y_1}} \right)}^2}} \\
\text{We have} \\
\Rightarrow \left( {{x_1},{y_1}} \right) = (3,2)\,\,and\,\,D = 5\,\,put\,\,the\,\,value \\
\Rightarrow D = \sqrt {{{\left( {u - {x_1}} \right)}^2} + {{\left( {v - {y_1}} \right)}^2}} \\
\Rightarrow {D^2} = {\left( {u - {x_1}} \right)^2} + {\left( {v - {y_1}} \right)^2} \\
\Rightarrow {5^2} = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\text{now from the equation} 3x - 4y = 1 \text{satisfy the point (u,v) then} \\
\Rightarrow 3u - 4v = 1\, \\
\Rightarrow 3u = 1 + 4v\, \\
\Rightarrow u = \dfrac{{1 + 4v}}{3} \]
\[\text{Now put the value into} \\
\Rightarrow 25 = {\left( {u - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{1 + 4v}}{3}\, - 3} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{1 + 4v - 9}}{3}} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = {\left( {\dfrac{{4v - 8}}{3}} \right)^2} + {\left( {v - 2} \right)^2} \\
\Rightarrow 25 = \left( {\dfrac{{16{v^2} + 64 - 64v}}{9}} \right) + \left( {{v^2} + 4 - 4v} \right) \\
\Rightarrow 25 \times 9 = \left( {16{v^2} + 64 - 64v} \right) + \left( {9{v^2} + 36 - 36v} \right) \\
\Rightarrow 225 = \left( {25{v^2} + 100 - 100v} \right) \\
\Rightarrow 25{v^2} + 100 - 100v = 225 \\
\Rightarrow 25{v^2} + 100 - 225 - 100v = 0 \\
\Rightarrow 25{v^2} - 125 - 100v = 0 \\
\Rightarrow 25\left( {{v^2} - 5 - 4v} \right) = 0 \\
\Rightarrow {v^2} - 4v - 5 = 0 \\
\Rightarrow {v^2} - 5v + v - 5 = 0 \\
\Rightarrow v\left( {v - 5} \right) + 1\left( {v - 5} \right) = 0 \\
\Rightarrow \left( {v + 1} \right)\left( {v - 5} \right) = 0 \\
or \\
\Rightarrow v = - 1\,\,or\,\,v = 5 \\
\text{now put it into the} \\
\Rightarrow u = \dfrac{{1 + 4v}}{3}\, \\
\text{then},v = - 1\,\, \\
\Rightarrow u = \dfrac{{1 + 4\left( { - 1\,} \right)}}{3}\, = \dfrac{{1 - 4}}{3}\, = \dfrac{{ - 3}}{3} = - 1 \\
u = - 1 \\
\text{ similarly for}\,v = 5 \\
\Rightarrow u = \dfrac{{1 + 4v}}{3} \\
then,v = 5\,\, \\
\Rightarrow u = \dfrac{{1 + 4\left( {5\,} \right)}}{3}\, = \dfrac{{1 + 20}}{3}\, = \dfrac{{21}}{3} = 7 \\
\Rightarrow u = 7 \\ \]
Now we got the pair of the points $\left( {u,v} \right) = \left( {7,5} \right)$ or $\left( {u,v} \right) = \left( { - 1, - 1} \right) $
Which is on the given line and our desired answer or point (7, 5), or (-1,-1)
Note: The straight line question is very straight forward we have just followed the instruction given by the question and we got the points on the line which are (7, 5), or (-1,-1), and both points on line having the distance from the point (3, 2) are 5 unite and can be cross-checked with the help of the distance formula
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
Which of the following Chief Justice of India has acted class 10 social science CBSE

Green glands are excretory organs of A Crustaceans class 11 biology CBSE

What if photosynthesis does not occur in plants class 11 biology CBSE

What is 1 divided by 0 class 8 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

10 slogans on organ donation class 8 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE
