Solve all your doubts with Teachoo Black (new monthly pack available now!)

Ex 10.3

Ex 10.3, 1 (i)

Ex 10.3, 1 (ii) Important

Ex 10.3, 1 (iii)

Ex 10.3, 2 (i)

Ex 10.3, 2 (ii)

Ex 10.3, 2 (iii) Important

Ex 10.3, 3 (i) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (ii) Deleted for CBSE Board 2023 Exams

Ex 10.3, 3 (iii) Important Deleted for CBSE Board 2023 Exams

Ex 10.3, 4

Ex 10.3, 5 Important

Ex 10.3, 6 (i) Important

Ex 10.3, 6 (ii)

Ex 10.3, 7

Ex 10.3, 8 Important You are here

Ex 10.3, 9 Important

Ex 10.3, 10

Ex 10.3, 11

Ex 10.3, 12 Important

Ex 10.3, 13

Ex 10.3, 14 Important

Ex 10.3, 15

Ex 10.3, 16 Important

Ex 10.3, 17 Important

Ex 10.3, 18 Important

Last updated at Feb. 3, 2020 by Teachoo

Ex 10.3, 8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. Let equation of line AB be x – 7y + 5 = 0 Let line CD be perpendicular to line AB and having x-intercept 3 Since Line CD has x-intercept 3 So, line CD passes through the point (3, 0) We have to find equation of line CD, Finding slope of line AB x − 7y + 5 = 0 − 7y = −x − 5 − 7y = −(x + 5) 7y = (x + 5) y = 1/7 (x + 5) y = 𝑥/7 + 5/7 The above equation is of the form y = mx + c where m = slope of line Thus, slope of line AB = 1/7 Now, Given that line AB and line CD are perpendicular We know that, product of slope of perpendicular lines is –1 So, (Slope of line AB) × (Slope of line CD) = –1 1/7 × (Slope of line CD) = –1 Slope of line CD = 7 × –1 Slope of line CD = –7 Equation of a line passing through a point (x0, y0) & having slope m is (y – y0)= m(x – x0) Equation of line CD passing through point P(3, 0) & having slope −7 is (y − 0) = –7(x – 3) y = –7 (x – 3) y = –7x – (–7) 3 y = –7x + 21 y + 7x = 21 Thus, the required equation of line is y + 7x = 21