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.

Related Further Mathematics GCSE answers

All answers ▸

In a chess club there are x boys and y girls. If ten more boys join and one more girl joins, there is an equal amount of boys and girls. Knowing that y = 2x+2, Calculate x and y. [4 marks]


Find the stationary point of 3x^2+7x


Work out the equation of the tangent to the curve y=x^2+5x-8 at the point (2,6)


A curve is mapped by the equation y = 3x^3 + ax^2 + bx, where a is a constant. The value of dy/dx at x = 2 is double that of dy/dx at x = 1. A turning point occurs when x = -1. Find the values of a and b.


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