Write an equation for each line described. Give answers in standard form for Exercises 11–20 and in slope-intercept form (if possible) for Exercises 21–32. through (-1,3), and (3,4)

Question

Accepted Answer

Welcome back. I am so glad you're here. We are asked to find the equation of a line passing through the points negative 35 and 76 and were asked to write the equation in standard form will be recalled from pre these lessons. That standard form is going to be A X plus B Y equals C or in other words, the X and Y terms together on one side of the equal sign and then the constant terms together on the opposite side of the equal sign. So how do we do that? Well, we're given two points. And so we're going to need to use those two points in order to get us to this standard form. And we can use those two points by using the point slope form. We want Y minus Y one equals M times X minus X one. Well, that uses one of the points. Where's the other point coming from? Well, that's in this M because we don't know the M, the M is the slope. And we find the slope using two points. We will use the formula Y two minus Y one divided by X two minus X one to find the slope. Then we'll plug that in to our point slope format, plug in one of the points and then we'll simplify it so that it looks like standard form quite a few steps here. So let's start working on finding the slope. We can write down our two points. We have negative 35 and and we can label our X ones and Y ones, etcetera. So this first one negative 35, let's label that can be our X one and Ry one And then our 76 that can be our X two and our Y two. And now we can plug the appropriate numbers back into the slope formula. So Y two is six minus Y one is five over X two is seven minus X one is negative three. Now, simplifying in the numerator six minus five is one in the denominator seven minus negative three. Well, that's the same as seven plus three and seven plus three is 10. So our slope is 1/ and now we can plug other things in here. We can bring down our Y minus and ry one. We said that was five and this equals 1/ times, bring down our X minus R X one is negative three. And now we can simplify in the parentheses, X minus negative three. That's the same as X plus three still in parentheses times 1/10 equal to Y minus five. Now, for our next step, we can get rid of the fraction by multiplying both sides times 10 by the denominator of that fraction. And so simplifying on the left, we have 10 times, why is 10, Y minus five times 10 is 50 equals on the right, the 10 times the 1/10 cancels out and we're left with X plus three. Now, recall, we want to get all of our exes and wise together on one side of the equation. And so let's go to where the X is positive. Let's move the Y to the right hand side. So we can subtract 10 Y from both sides of the equation. And then we want to move the constant terms to the other side away from the X and Y. So let's subtract three from both sides. And then on the left 10, Y -10, Y is nothing negative 50 minus three is negative 53. And that equals on the right. We just still have that X and then minus 10 Y and then those threes cancel out. So this is our equation in standard form. We have X minus 10, Y equals a negative 53. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

Write an equation for each line described. Give answers in standard form for Exercises 11–20 and in slope-intercept form (if possible) for Exercises 21–32. through (-1,3), and (3,4)

Verified Solution

Key Concepts

Slope-Intercept Form

Standard Form of a Linear Equation

Finding the Slope Between Two Points