SOMEONE HELP MEEEEEE 75 POINTS TO THE PERSON THAT HELPS1. Indicate the equation of the given line in standard form.The line with slope 9/7 and containing the midpoint of the segment whose endpoints are (2, -3) and (-6, 5).2. Indicate the equation of the given line in standard form.The line through the midpoint of and perpendicular to the segment joining points (1, 0) and (5, -2).3. Indicate the equation of the given line in standard form.The line containing the midpoints of the legs of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices.4. Indicate the equation of the given line in standard form.The line containing the hypotenuse of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices.5. Indicate the equation of the given line in standard form.The line containing the longer diagonal of a quadrilateral whose vertices are A (2, 2), B(-2, -2), C(1, -1), and D(6, 4).6. Indicate the equation of the given line in standard form.The line containing the median of the trapezoid whose vertices are R(-1, 5) , S(1, 8), T(7, -2), and U(2, 0).7. Indicate the equation of the given line in standard form.The line containing the altitude to the hypotenuse of a right triangle whose vertices are P(-1, 1), Q(3, 5), and R(5, -5).8. Indicate the equation of the given line in standard form.The line containing the diagonal of a square whose vertices are A(-3, 3), B(3, 3), C(3, -3), and D(-3, -3). Find two equations, one for each diagonal.

Answer:Part 1) [tex]9x-7y=-25[/tex]Part 2) [tex]2x-y=2[/tex]Part 3) [tex]x+8y=22[/tex]  Part 4) [tex]x+8y=35[/tex]Part 5) [tex]3x-4y=2[/tex]Part 6) [tex]10x+6y=39[/tex]Part 7) [tex]x-5y=-6[/tex]Part 8) case A) The equation of the diagonal AC is [tex]x+y=0[/tex]case B) The equation of the diagonal BD is [tex]x-y=0[/tex]Step-by-step explanation:Part 1) step 1Find the midpointThe formula to calculate the midpoint between two points is equal to[tex]M=(\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]substitute the values[tex]M=(\frac{2-6}{2},\frac{-3+5}{2})[/tex][tex]M=(-2,1)[/tex]step 2The equation of the line into point slope form is equal to[tex]y-1=\frac{9}{7}(x+2)\\ \\y=\frac{9}{7}x+\frac{18}{7}+1\\ \\y=\frac{9}{7}x+\frac{25}{7}[/tex]step 3Convert to standard formRemember that the equation of the line into standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integer, and B, and C are integers[tex]y=\frac{9}{7}x+\frac{25}{7}[/tex]Multiply by 7 both sides[tex]7y=9x+25[/tex][tex]9x-7y=-25[/tex]Part 2) step 1Find the midpointThe formula to calculate the midpoint between two points is equal to[tex]M=(\frac{x1+x2}{2},\frac{y1+y2}{2})[/tex]substitute the values[tex]M=(\frac{1+5}{2},\frac{0-2}{2})[/tex][tex]M=(3,-1)[/tex]step 2Find the slopeThe slope between two points is equal to[tex]m=\frac{-2-0}{5-1}=-\frac{1}{2}[/tex]step 3we know thatIf two lines are perpendicular, then the product of their slopes is equal to -1Find the slope of the line perpendicular to the segment joining the given points[tex]m1=-\frac{1}{2}[/tex][tex]m1*m2=-1[/tex]therefore[tex]m2=2[/tex]step 4The equation of the line into point slope form is equal to[tex]y-y1=m(x-x1)[/tex]we have[tex]m=2[/tex] and point [tex](1,0)[/tex][tex]y-0=2(x-1)\\ \\y=2x-2[/tex]step 5Convert to standard formRemember that the equation of the line into standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integer, and B, and C are integers[tex]y=2x-2[/tex][tex]2x-y=2[/tex]Part 3) In this problem AB and BC are the legs of the right triangle (plot the figure)step 1Find the midpoint AB[tex]M1=(\frac{-5+1}{2},\frac{5+1}{2})[/tex][tex]M1=(-2,3)[/tex]step 2Find the midpoint BC[tex]M2=(\frac{1+3}{2},\frac{1+4}{2})[/tex][tex]M2=(2,2.5)[/tex]step 3Find the slope M1M2The slope between two points is equal to[tex]m=\frac{2.5-3}{2+2}=-\frac{1}{8}[/tex]step 4The equation of the line into point slope form is equal to[tex]y-y1=m(x-x1)[/tex]we have[tex]m=-\frac{1}{8}[/tex] and point [tex](-2,3)[/tex][tex]y-3=-\frac{1}{8}(x+2)\\ \\y=-\frac{1}{8}x-\frac{1}{4}+3\\ \\y=-\frac{1}{8}x+\frac{11}{4}[/tex]step 5Convert to standard formRemember that the equation of the line into standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integer, and B, and C are integers[tex]y=-\frac{1}{8}x+\frac{11}{4}[/tex]Multiply by 8 both sides[tex]8y=-x+22[/tex] [tex]x+8y=22[/tex]  Part 4) In this problem the hypotenuse is AC (plot the figure)step 1Find the slope ACThe slope between two points is equal to[tex]m=\frac{4-5}{3+5}=-\frac{1}{8}[/tex]step 2The equation of the line into point slope form is equal to[tex]y-y1=m(x-x1)[/tex]we have[tex]m=-\frac{1}{8}[/tex] and point [tex](3,4)[/tex][tex]y-4=-\frac{1}{8}(x-3)[/tex][tex]y=-\frac{1}{8}x+\frac{3}{8}+4[/tex][tex]y=-\frac{1}{8}x+\frac{35}{8}[/tex]step 3Convert to standard formRemember that the equation of the line into standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integer, and B, and C are integers[tex]y=-\frac{1}{8}x+\frac{35}{8}[/tex]Multiply by 8 both sides[tex]8y=-x+35[/tex][tex]x+8y=35[/tex]Part 5)   The longer diagonal is the segment BD (plot the figure)   step 1Find the slope BDThe slope between two points is equal to[tex]m=\frac{4+2}{6+2}=\frac{3}{4}[/tex] step 2The equation of the line into point slope form is equal to[tex]y-y1=m(x-x1)[/tex]we have[tex]m=\frac{3}{4}[/tex] and point [tex](-2,-2)[/tex][tex]y+2=\frac{3}{4}(x+2)[/tex][tex]y=\frac{3}{4}x+\frac{6}{4}-2[/tex][tex]y=\frac{3}{4}x-\frac{2}{4}[/tex]step 3Convert to standard formRemember that the equation of the line into standard form is equal to[tex]Ax+By=C[/tex]whereA is a positive integer, and B, and C are integers[tex]y=\frac{3}{4}x-\frac{2}{4}[/tex]Multiply by 4 both sides[tex]4y=3x-2[/tex][tex]3x-4y=2[/tex]Note The complete answers in the attached file