4 Find the coordinates of the point on the tive presiented by \( 2 x-3 y+7=0 \) which is equatistant from the point \( (-4,-8) \) and \( (7,-1) \)
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To find the point that is equidistant from the points \((-4, -8)\) and \((7, -1)\), we first need to locate the midpoint of these two points. The midpoint \(M\) is calculated as \(M\left(\frac{-4 + 7}{2}, \frac{-8 - 1}{2}\right) = M\left(\frac{3}{2}, -\frac{9}{2}\right)\). Next, we observe that the slope of the line segment connecting the two points is given by \(m = \frac{-1 + 8}{7 + 4} = \frac{7}{11}\). The line that is equidistant from both points can be represented by its perpendicular bisector. The slope of the perpendicular bisector is the negative reciprocal of the slope of the line segment, which is \(-\frac{11}{7}\). Using the midpoint \(\left(\frac{3}{2}, -\frac{9}{2}\right)\) and the slope \(-\frac{11}{7}\), the equation of the perpendicular bisector can be formulated. Now, let’s solve for where this intersects the line given by \(2x - 3y + 7 = 0\). Beginning with that equation, we can express \(y\) in terms of \(x\) and derive where this intersects the perpendicular line. Finally, we substitute and solve for the relevant coordinates. The point where these two lines intersect will yield the desired equidistant point!