Question

which point is on the line that passes through point z and is perpendicular to line ab? (-4,1) (1,-2) (2,0) (4,4)

Explanation:

Step1: Find the slope of line \(AB\)

The coordinates of \(A(-3,4)\) and \(B(0, - 3)\).
The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
So, \(m_{AB}=\frac{-3 - 4}{0-(-3)}=\frac{-7}{3}\).

Step2: Find the slope of the perpendicular line

If two lines are perpendicular, \(m_1\times m_2=-1\). Let \(m_1 = m_{AB}=-\frac{7}{3}\), then \(m_2=\frac{3}{7}\).
The equation of the line passing through \(Z(0,2)\) (from the graph \(Z\) has coordinates \((0,2)\)) using the point - slope form \(y - y_1=m(x - x_1)\) is \(y-2=\frac{3}{7}(x - 0)\), so \(y=\frac{3}{7}x+2\).

Step3: Check each point

• For \((-4,1)\): \(y=\frac{3}{7}x + 2\), substitute \(x=-4\), \(y=\frac{3}{7}\times(-4)+2=\frac{-12 + 14}{7}=\frac{2}{7}

eq1\).

• For \((1,-2)\): substitute \(x = 1\), \(y=\frac{3}{7}\times1+2=\frac{3 + 14}{7}=\frac{17}{7}

eq-2\).

• For \((2,0)\): substitute \(x = 2\), \(y=\frac{3}{7}\times2+2=\frac{6+14}{7}=\frac{20}{7}

eq0\).

• For \((4,4)\): substitute \(x = 4\), \(y=\frac{3}{7}\times4+2=\frac{12 + 14}{7}=4\).

Answer:

\((4,4)\)