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in 20 and 21, use the coordinate plane at the right. the graph shows th…

Question

in 20 and 21, use the coordinate plane at the right. the graph shows the locations of point g and point h. point j is graphed at (n, -3). the distance from point h to point j is equal to the distance from point h to point g. 20. what is the distance from point h to point j? 21. what is the value of n?

Explanation:

Step1: Calculate the distance from \(H(n,-6)\) to \(J(n,-3)\)

Since the \(x -\)coordinates of \(H\) and \(J\) are the same (\(x = n\)), we use the distance formula for two points \((x_1,y_1)\) and \((x_2,y_2)\) when \(x_1=x_2\), \(d=\vert y_2 - y_1\vert\).
Here, \(x_1=x_2=n\), \(y_1=-6\), \(y_2 = - 3\). Then \(d=\vert-3-(-6)\vert=\vert-3 + 6\vert=\vert3\vert = 3\)

Step2: Find the \(x -\)coordinate of \(G\)

Since \(H(n,-6)\) and \(G\) have the same \(y -\)coordinate (\(y=-6\)) and the distance from \(H\) to \(G\) is equal to the distance from \(H\) to \(J\) (which is \(3\)).
If we assume \(G\) has coordinates \((m,-6)\) and \(H(n,-6)\), and the distance between them is \(3\). Looking at the coordinate - plane (assuming \(G\) is \(3\) units to the left of \(H\) in the \(x -\)direction), if we assume \(G\) has \(x -\)coordinate \(n-3\) (by the distance formula \(d=\vert x_2 - x_1\vert\) with \(y_1=y_2=-6\)). But since \(J(n,-3)\) and \(H(n,-6)\) have the distance \(3\) and \(HG = HJ\), and if we assume \(G\) is symmetric in some way (using the fact that for vertical distance \(HJ = 3\) and horizontal distance \(HG=3\)).
Since \(H(n,-6)\) and \(G\) have \(y=-6\) and distance \(3\), if we assume \(G\) is \(3\) units to the left of \(H\) (assuming the coordinate - plane's left - right orientation), then \(n-3\) (for \(G\)'s \(x -\)coordinate). But if we consider the fact that for two points \((x_1,y_1)\) and \((x_2,y_2)\) with \(y_1 = y_2\), \(d=\vert x_2 - x_1\vert\). Since \(HJ=HG = 3\) and \(J(n,-3)\), \(H(n,-6)\), for \(G\) (assuming \(G\) has \(y=-6\) and \(HG = 3\)), if we assume \(G\) is at \((n - 3,-6)\) (using the horizontal distance formula \(d=\vert x_2 - x_1\vert\) with \(y_1=y_2=-6\)). But if we consider the coordinate - plane's structure (assuming \(G\) is at \((-3,-6)\) and \(H\) is at \((0,-6)\) (by visual inspection of the coordinate - plane's general form where the \(y -\)axis is \(x = 0\)). So \(n=0\)

Answer:

  1. The distance from point \(H\) to point \(J\) is \(3\)
  2. The value of \(n\) is \(0\)