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Prove that the line between \((1,-1)\) and \((4,-3)\) is perpendicular to the line \(3x-2y-4=0\) (2 marks) --- 6 WORK AREA LINES (style=lined) ---
\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)
\(\text {Line 1 gradient:}\)
\(P_1 (1,-1), P_2(4,-3) \)
\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
\(\text {Line 2 gradient:}\)
\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)
\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)
\(\therefore\ \text{Lines are perpendicular.}\)
\(\text {Perpendicular lines}\ \ \Rightarrow\ m_1 \times m_2 = -1\)
\(\text {Line 1 gradient:}\)
\(P_1 (1,-1), P_2(4,-3) \)
\(m_1=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{-3+1}{4-1}=-\dfrac{2}{3}\)
\(\text {Line 2 gradient:}\)
\(3x-2y-4=0\ \ \Rightarrow \ y= \dfrac{3}{2}x-2\ \ \Rightarrow m_2=\dfrac{3}{2}\)
\(m_1 \times m_2 = -\dfrac{2}{3} \times \dfrac{3}{2} = -1\)
\(\therefore\ \text{Lines are perpendicular.}\)
What is the gradient of the line `2x + 3y + 4 = 0`?
`A`
| `2x + 3y + 4` | `= 0` |
| `3y` | `= -2x-4` |
| `y` | `= -2/3 x-4/3` |
`=> A`
What is the slope of the line with equation `2x-4y + 3 = 0`?
`C`
| `2x-4y + 3` | `= 0` |
| `4y` | `= 2x + 3` |
| `y` | `= 1/2 x + 3/4` |
`=> C`