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

Related Further Mathematics GCSE answers

All answers ▸

The equation of the line L1 is y = 3x – 2 The equation of the line L2 is 3y – 9x + 5 = 0 Show that these two lines are parallel.


The circle c has equation x^2+ y ^2=1 . The line l has gradient 3 and intercepts the y axis at the point (0, 1). c and l intersect at two points. Find the co-ordinates of these points.


x^3 + 2x^2 - 9x - 18 = (x^2 - a^2)(x + b) where a,b are integers. Work out the three linear factors of x^3 + 2x^2 - 9x - 18. (Note: x^3 indicates x cubed and x^2 indicates x squared).


How would I solve the following equation d^2x/dt^2 + 5dx/dt + 6x = 0