The point P has coordinates (3, 4) The point Q has coordinates (a, b) A line perpendicular to PQ is given by the equation 3x + 2y = 7 Find an expression for b in terms of a

Perpendicular gradients multiply to give -1. The gradient of the perpendicular line (y=-3/2x+7/2) is -3/2 , so the gradient of PQ is 2/3. Using gradient formula change in y/ change in x :(4-b)/(3-a) = 2/3 Simplifying:4-b=(6-2a)/3 12-3b=6-2a 3b=6+2a In terms of b: b=2+2/3a

Answered by Lizzie D. Maths tutor

2686 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

P is a point on the circle with equation x^2 + y^2 = 80. P has x-coordinate 4 and is below the x-axis.Work out the equation of the tangent to the circle at P.


How do I complete the square?


Simplify, leaving your answer as a quadratic: (2x + 3)/(x+4) - (3x - 6) = 4


Solve x^2 + x - 2 = 0


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences