Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form? 7x + 2y = –1
7x + 2y = 1
14x + 4y = –1
14x + 4y = 1

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  (b) 7x + 2y = 1

Step-by-step explanation:

You don't need to know how to find the equation. You just need to know how to determine if a point satisfies the equation. Try one of the points and see which equation fits. (The numbers are smaller for point K, so we prefer to use that one.)

  7(1) +2(-3) = 1 ≠ -1 . . . . . tells you choice A doesn't work, and choice B does

The equation is ...

  7x +2y = 1

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Additional comment

The equations of choices C and D have coefficients with a common factor of 2. If the constant also had a factor of 2, we could say these equations are not in standard form, and we could reject them right away. Since the two points have integer values for x and y, we can reject these equations anyway: the sum of even numbers cannot be odd.

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

the first one is x^2+2x+2

the second one is x^2+4

the last one is x^2-4x+8