Gradient and y-intercept questions with answers

Questions

1. Question 1From y = mx + c

   Find the gradient of the line $y = -6x + 5$.

   Show the worked solution

   1. Step 1: The gradient m is the number in front of x$m = -6$
2. Question 2From ax + by = c

   Find the gradient of the line $-x + 2y = 5$.

   Show the worked solution

   1. Step 1: Make y the subject: move the x term across (it changes sign)$2y = x + 5$
   2. Step 2: Divide every term by the number in front of y (dividing by a negative flips every sign)$y = \dfrac{1}{2}x + \dfrac{5}{2}$
   3. Step 3: Compare with y = mx + c: the gradient is the number in front of x$m = \dfrac{1}{2}$
3. Question 3From two points on a line

   The line passes through the points $A$ and $B$ shown. Find its gradient.

   Show the worked solution

   1. Step 1: Read A and B off the grid. Gradient = rise ÷ run, from A to B (the rise is negative if the line goes down)$A(-3, -2)$ and $B(1, 4)$Rise: $4 - (-2) = 6$Run: $1 - (-3) = 4$
   2. Step 2: Divide and simplify$m = \dfrac{6}{4}$$m = \dfrac{3}{2}$
4. Question 4Positive, negative, zero or undefined

   State whether the gradient of the line $y = -4$ is positive, negative, zero or not defined.

   Show the worked solution

   1. Step 1: Picture the line y = -4: every point on it has the same y-coordinateThe line goes flat (horizontal).
   2. Step 2: Up: positive. Down: negative. Flat: zero. Vertical: not defined (the run is zero)So the gradient is zero.
5. Question 5The equation of a line through two points

   A straight line passes through the points $A(3, 2)$ and $B(-1, -6)$. Find the equation of the line, in the form $y = mx + c$.

   Show the worked solution

   1. Step 1: Gradient = rise ÷ run, from A to B$m = \dfrac{-6 - 2}{-1 - 3}$$\phantom{m} = \dfrac{-8}{-4}$$\phantom{m} = 2$
   2. Step 2: Put the gradient and the point A(3, 2) into y = mx + c, then solve for c$2 = 2 \times 3 + c$$c = -4$
   3. Step 3: Write the equation$y = 2x - 4$
6. Question 6Where a line meets the axes; the triangle it makes

   The line $10x - 3y = -30$ meets the $x$-axis at $P$ and the $y$-axis at $Q$. $O$ is the origin. Find the area of triangle $OPQ$.

   Show the worked solution

   1. Step 1: P is on the x-axis: put y = 0 and solve$10x = -30$$x = -3$
   2. Step 2: Q is on the y-axis: put x = 0 and solve$-3y = -30$$y = 10$
   3. Step 3: Triangle OPQ has a right angle at O: area = ½ × OP × OQ (lengths are positive)$\dfrac{1}{2} \times 3 \times 10 = 15$ square units