Find the Distance of the Point (4, 5) from the Straight Line 3x − 5y + 7 = 0. - Mathematics

Find the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0.

Solution

Comparing ax + by + c = 0 and 3x − 5y + 7 = 0, we get:
a = 3, b = − 5 and c = 7
So, the distance of the point (4, 5) from the straight line 3x − 5y + 7 = 0 is

$d = \left| \frac{a x_1 + b y_1 + c}{\sqrt{a^2 + b^2}} \right|$

$\Rightarrow d = \left| \frac{3 \times 4 - 5 \times 5 + 7}{\sqrt{3^2 + \left( - 5 \right)^2}} \right| = \frac{6}{\sqrt{34}}$

Hence, the required distance is $\frac{6}{\sqrt{34}}$.

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.15 | Q 1 | Page 107