NCERT Class XI Mathematics - Straight Lines - Solutions
© examsnet.com
Question : 24
Total: 74
Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Solution:
Let M(2, 5) and N(–3, 6) be the end points of the given line segment.
Slope of MN =
Since LP ⊥ MN
∴ Slope of LP is 5 and passing through (–3, 5)
Equation of line LP is (y – 5) = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0.
Slope of MN =
Since LP ⊥ MN
∴ Slope of LP is 5 and passing through (–3, 5)
Equation of line LP is (y – 5) = 5(x + 3)
⇒ y – 5 = 5x + 15
⇒ 5x – y + 20 = 0.
© examsnet.com
Go to Question: