QUESTION IMAGE
Question
which point could be on the line that is perpendicular to mn and passes through point k? (4,8) (0,-12) (2,2) (5,13)
Step1: Find the slope of line \(MN\)
The coordinates of \(M(3,3)\) and \(N(-2,2)\).
The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
So, \(m_{MN}=\frac{3 - 2}{3-(-2)}=\frac{1}{5}\).
Step2: Find the slope of the perpendicular line
If two lines are perpendicular, the product of their slopes is \(- 1\). Let the slope of the perpendicular line be \(m\).
Since \(m_{MN}\times m=-1\) and \(m_{MN}=\frac{1}{5}\), then \(m=-5\).
Step3: Assume the equation of the line passing through \(K(3,-3)\)
The point - slope form of a line is \(y - y_1=m(x - x_1)\). Here \(x_1 = 3,y_1=-3,m=-5\).
So, \(y+3=-5(x - 3)\), which simplifies to \(y=-5x + 12\).
Step4: Check each point
- For \((4,8)\): Substitute \(x = 4\) into \(y=-5x + 12\), \(y=-5\times4+12=-20 + 12=-8
eq8\).
- For \((0,-12)\): Substitute \(x = 0\) into \(y=-5x + 12\), \(y=-5\times0+12 = 12
eq-12\).
- For \((2,2)\): Substitute \(x = 2\) into \(y=-5x + 12\), \(y=-5\times2+12=-10 + 12=2\).
- For \((5,13)\): Substitute \(x = 5\) into \(y=-5x + 12\), \(y=-5\times5+12=-25 + 12=-13
eq13\).
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\((2,2)\)