If Qh-k is the image of the point Px1-y1 w-r-t- the straight line ax-by-c-0- Then h - x1-a - k - y1-b - - 2 ax1 - by1 - c-a2 - b2 or h - x1-a - k - y1-b - - 2 ax1 - by1 - c-a2 - b2 and find the image of 1-2 w-r-t- the straight line 2x-3y-5-0-

[ h 1/2 = k + 2/ 3 = 2( ( 2 × 1) 3( 2) + 5)/( 2 )^2 + ( 3)^2; h 1/2 = k + 2/ 3 = 2( 13)/13 = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Differentiability in an Interval

If Qh,k is ...

Question

If $Q (h, k)$ is the image of the point $P (x_{1}, y_{1})$ w.r.t. the straight line $a x + b y + c = 0$ . Then $(h - x_{1}) : a = (k - y_{1}) : b = - 2 (a x_{1} + b y_{1} + c) : a^{2} + b^{2}$ (or) $\frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{- 2 (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$ and find the image of $(1, - 2)$ w.r.t. the straight line $2 x - 3 y + 5 = 0$ .

Open in App

Solution

$\begin{matrix} \frac{h - 1}{2} = \frac{k + 2}{- 3} = \frac{- 2 ((2 \times 1) - 3 (- 2) + 5)}{{(2)}^{2} + {(- 3)}^{2}} \frac{h - 1}{2} = \frac{k + 2}{- 3} = \frac{- 2 (13)}{13} = - 2 \frac{h - 1}{2} = - 2 h = - 3 \frac{k + 2}{- 3} = - 2 k = - 8 ∴ p o i n t \equiv (- 3, - 8) \end{matrix}$

Suggest Corrections

Similar questions

Find the length of the perpendicular from the point $(x^{1}, y^{1})$ to the straight line Ax + By + C = 0, the axes being inclined at an angle $ω$ , and the equation being written such that C is a negative quantity.

Q. Assertion :If

a, b, c

are in A. P. then straight line

a x + b y + c = 0

will always passes through a fixed point

(1, - 2)

. Reason: A straight line

a x + b y + c = 0

will pass through a point

(x_{1}, y_{1})

a x_{1} + b y_{1} + c = 0

Q. Let t =

| \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} |

. Find the lenght of perpendicular from

p (x_{1}, y_{1})

on ax+by+c=0 in terms of t.

Two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ lie on the same side of the straight line ax + by +c = 0 if ( $a x_{1} + b y_{1} + c$ ) ( $a x_{2} + b y_{2} + c$ ) is positive

Let t = $| \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} |$ . Find the length of perpendicular from $p (x_{1}, y_{1})$ on ax+by+c=0 in terms of t.

Join BYJU'S Learning Program

If Q(h,k) is the image of the point P(x1,y1) w.r.t. the straight line ax+by+c=0. Then (h−x1):a=(k−y1):b=−2(ax1+by1+c):a2+b2 (or) h−x1a=k−y1b=−2(ax1+by1+c)a2+b2 and find the image of (1,−2) w.r.t. the straight line 2x−3y+5=0.

h−12=k+2−3=−2((2×1)−3(−2)+5)(2)2+(−3)2h−12=k+2−3=−2(13)13=−2h−12=−2h=−3k+2−3=−2k=−8∴point≡(−3,−8)

$\begin{matrix} \frac{h - 1}{2} = \frac{k + 2}{- 3} = \frac{- 2 ((2 \times 1) - 3 (- 2) + 5)}{{(2)}^{2} + {(- 3)}^{2}} \frac{h - 1}{2} = \frac{k + 2}{- 3} = \frac{- 2 (13)}{13} = - 2 \frac{h - 1}{2} = - 2 h = - 3 \frac{k + 2}{- 3} = - 2 k = - 8 ∴ p o i n t \equiv (- 3, - 8) \end{matrix}$