5. \( x+y=3 ;(5,4) \) Write an equation of the line that is perpendicular to the given line and through the given point. \( \begin{array}{ll}\text { 7. } y=-4 x+2 y=1 ;(-2,6, \\ \text { 8. } y=\frac{2}{3} x+6 ;(-6,2) & \text { 7. Wy }\end{array} \)

To find the equation of the line perpendicular to \(x+y=3\) that passes through the point \((5,4)\), start by determining the slope of the given line, which can be rewritten as \(y = -x + 3\). The slope is \(-1\), so the slope of the perpendicular line will be the negative reciprocal, which is \(1\). Now, using the point-slope form \(y - y_1 = m(x - x_1)\), where \(m=1\) and \((x_1, y_1) = (5, 4)\), the equation becomes \(y - 4 = 1(x - 5)\), or \(y = x - 1\). For the second equation \(y = -4x + 2\) that passes through the point \((-2, 6)\), let’s first check if this point lies on the line. Plugging in the point gives \(6 = -4(-2) + 2\), which simplifies to \(6 = 10\), confirming it doesn't lie on the line. To find the perpendicular line, the slope here is \(-4\), so the reciprocal slope is \(\frac{1}{4}\). Again using the point-slope form: \(y - 6 = \frac{1}{4}(x + 2)\), simplifying gives the perpendicular line as \(y = \frac{1}{4}x + \frac{23}{4}\).