A straight line passes trough the points A(-4;7); B(6;-5); C(8;t). Use an algebraic method to work out the value of t.

We can write the equation of a straight line like y=ax+b, where a and b are constants. If a point is on the line, the equation will be true for the x,y of the point.Now start with point A: 7 = (-4)a + b. Resolve this equation to b. Add (-4)a both side: 7+4a=b. Now see point B. (-5) = 6a+b. We know, that b=7+4a, so (-5) = 6a + 7 + 4a = 10a + 7. Substract 7 from both sides: (-12) = 10a, so divide by 10, and we get that -1.2 = a. From this we can calculate b: b = 7 + 4•(-1.2) = 7 - 4.8 = 2.2 = b.Now we need to resolve the equation t = 8a + b = 8•(-1.2) + 2.2 = (-9.6) + 2.2 = 7.4.So the result is t = 7.4.

MF
Answered by Márk F. Further Mathematics tutor

6689 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Given that xy=2 and y=3x+5, find x and y. Do not use trial and improvement.


In the expansion of (x-7)(3x**2+kx-3) the coefficient of x**2 is 0. i) Find the value of k ii) Find the coefficient of x. iii) write the fully expanded equation in terms of x


Find the coordinates of the minimum point of the function y=(x-5)(2x-2)


Consider the Matrix M (below). Find the determiannt of the matrix M by using; (a) cofactor expansion along the first row, (b) cofactor expansion along the second column


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences