Answer:
The third choice is the one you want
Step-by-step explanation:
If we are to write the equation of a line perpendicular to WX, we first must determine what the slope of the WX is, because the line perpendicular to WX has a slope that is the flip of the slope of WX with the opposite sign. Solving for y takes care of finding the slope of WX:
2x + y = -5 so
y = -2x - 5
The slope is -2. That means that the reciprocal slope is 1/2. Using that slope along with the coordinates x = -1 and y = -2, we first write the line using point-slope form and then solve it for y. Start by filling in the m, the x value and the y value:

Getting rid of the double negatives gives us:

Distributing then gives us:

And finally solving for y (I am going to express the 2 on the left as 4/2 when I move it by subtraction in order to add those fractions):

And the final equation in slope-intercept form is:

Answer:
1) x=3, y=-3
Step-by-step explanation:
10x + 7y = 9 —— (1)
-4x -7y = 9 —— (2)
(1) + (2)
6x = 18
x = 18/6
x = 3
put x=3 in (1)
10(3) + 7y = 9
30 + 7y = 9
7y = 9 - 30
7y = -21
y = -21/7
y = -3
please give me brainlist if it is helpful for you
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
