First slide

Equation of line in Straight lines

Question

The image of the line $x - y + 3 = 0$ on x-axis is $px + qy + r = 0$ then the value of $p + q + r =$

Moderate

A

2
B

3
C

-5
D

5

Solution

To get the image of any line $L_{1} = 0$ with respect to the other line $L_{2} = 0$ then find the equation of the line passing through images of any two points on the line $L_{1} = 0$ with respect to the line $L_{2} = 0$
Two points on the line $x - y + 3 = 0$ are $(0, 3)$ and $(1, 4)$
The image of a point $(a, b)$ on x-axis is $(a, - b)$
Hence the images of $(0, 3)$ and $(1, 4)$ on x-axis are $(0, - 3)$ and $(1, - 4)$
Equation of the line passing through the points $(0, - 3)$ and $(1, - 4)$ is
$\begin{matrix} y + 3 = \frac{- 4 + 3}{1 - 0} (x - 0) \\ y + 3 = - x \\ x + y + 3 = 0 \end{matrix}$
Compare this with $px + qy + r = 0$ ,
Hence $p = 1, q = 1, r = 3$ .
Therefore $p + q + r = 1 + 1 + 3 = 5$

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