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Find the point on the \[3x - 4y - 1{\text{ }} = 0\], where are at a distance of 5 units from the point (3, 2)

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Last updated date: 21st Jul 2024
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Answer
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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