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
8078 Views
See similar Further Mathematics GCSE tutorsRelated Further Mathematics GCSE answers
All answers ▸A circle has equation x^{2}-8x+y^{2}-6y=d. A line is tangent to this circle and passes through points A and B, (0,17) and (17,0) respectively. Find the radius of the circle.
Using differentiation, show that f(x) = 2x^3 - 12x^2 + 25x - 11 is an increasing function.
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.
y=(6x^9 +x^8)/(2x^4), work out the value of d^2y/dx^2 when x=0.5