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.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
t

Perpendicular distant of $p (x_{1}, y_{1})$ from $a x_{1} + b y_{1} + c$ is given by $| \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} |$ .
We can derive this formula in the following way.

We want to find d or PR. we can find d if we know the co-ordinates of R.
1. Find the slope of PR assuming $R \equiv (x_{2}, y_{2})$
2. Find the slope of PR as $\frac{- 1}{m}$ , where m is the slope of L. (PR and L are perpendicular)
3. Equate the slopes in 1 and 2 to get a relation between $x_{2} a n d y_{2}$
4. We get one more relation because $(x_{2}, y_{2})$ lies on L.
5. Use 3 and 4 to find $x_{2}, y_{2}$
6. Find the distance using distance formula
It is good to remember both formula and the method.

Suggest Corrections

Similar questions

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.

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.

Distance between two parallel lines Ax₁ + By₁ + C₁ = 0 and Ax₂ + By₂ + C₂ = 0 is:

Q. In

△ A B C, \frac{s i n (A - B)}{s i n (A + B)} =

[MP PET 1986]

Q. Let P = a² − b² + 2ab, Q = a² + 4b² − 6ab, R = b² + 6, S = a² − 4ab and T = −2a² + b² − ab + a. Find P + Q + R + S − T.

Join BYJU'S Learning Program

Let t = |ax1+by1+c√a2+b2|. Find the lenght of perpendicular from p(x1,y1) on ax+by+c=0 in terms of t.

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.