7.7.2 Coordinate Geometry Short Questions (Question 3, 4 & 5)

Question 3:

The equation of the straight lines CD and EF are 5x + y – 4 = 0 and

\frac{x}{7} - \frac{y}{h} = 1

$\frac{x}{7} - \frac{y}{h} = 1$ . If CD and EF are parallel, find the value of h.

Solution:

Two parallel lines have same gradient.

5x + y – 4 = 0

y = –5x + 4, m_CD= –5

For the straight line EF,

\begin{array}{l} \frac{x}{7} - \frac{y}{h} = 1 \\ m_{E F} = - (\frac{y -intercept}{x -intercept}) = - (\frac{- h}{7}) = \frac{h}{7} \\ m_{C D} = m_{E F} \\ - 5 = \frac{h}{7} \\ h = - 35 \end{array}

$\begin{array}{l} \frac{x}{7} - \frac{y}{h} = 1 \\ m_{E F} = - (\frac{y -intercept}{x -intercept}) = - (\frac{- h}{7}) = \frac{h}{7} \\ m_{C D} = m_{E F} \\ - 5 = \frac{h}{7} \\ h = - 35 \end{array}$

Question 4:

The straight line

\frac{x}{5} + \frac{y}{p} = 1

$\frac{x}{5} + \frac{y}{p} = 1$ has a y-intercept of 3 and is parallel to the straight line y + qx = 0. Determine the value of p and of q.

Solution:

\begin{array}{l} Given y-intercept of the straight line \frac{x}{5} + \frac{y}{p} = 1 is 3, \\ ∴ p = 3 \\ Gradient of the straight line = - \frac{3}{5} \\ For another straight line y + q x = 0, y = - q x \\ As two parallel lines have same gradient, \\ - q = - \frac{3}{5} \\ q = \frac{3}{5} \end{array}

$\begin{array}{l} Given y-intercept of the straight line \frac{x}{5} + \frac{y}{p} = 1 is 3, \\ ∴ p = 3 \\ Gradient of the straight line = - \frac{3}{5} \\ For another straight line y + q x = 0, y = - q x \\ As two parallel lines have same gradient, \\ - q = - \frac{3}{5} \\ q = \frac{3}{5} \end{array}$

Question 5:

The equations of two straight lines

\frac{y}{7} + \frac{x}{4} = 1

$\frac{y}{7} + \frac{x}{4} = 1$ y = 4x + 21. Determine whether the lines are perpendicular to each other.

Solution:

\begin{array}{l} For the straight line \frac{y}{7} + \frac{x}{4} = 1, gradient = - \frac{7}{4} \\ For the straight line 7 y = 4 x + 21, \\ y = \frac{4}{7} x + 3, gradient = \frac{4}{7} \\ - \frac{7}{4} \times \frac{4}{7} = - 1 \end{array}

$\begin{array}{l} For the straight line \frac{y}{7} + \frac{x}{4} = 1, gradient = - \frac{7}{4} \\ For the straight line 7 y = 4 x + 21, \\ y = \frac{4}{7} x + 3, gradient = \frac{4}{7} \\ - \frac{7}{4} \times \frac{4}{7} = - 1 \end{array}$

Hence, the two straight lines are perpendicular to each other.

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