Video: CBSE Class X • Pack 5 • 2014 • Question 7

CBSE Class X • Pack 5 • 2014 • Question 7

04:49

Video Transcript

If 𝐴: 𝑥, two; 𝐵: negative three, negative four; and 𝐶: seven, negative five are collinear, find the value of 𝑥.

So if points are collinear, that means they lie on the same line. So we could have something like this. And because they’re collinear, lying on the same line, the segments joining these points will have the same slope.

Slope 𝑚 is the change in 𝑦, which is the rise, divided by the change in 𝑥, the run. So if we found the slope of segment 𝐵𝐶, we could use it to find the equation of the line between all three points. And then once we found the equation of the line, we can plug in one of the values of 𝐴, which is two — we’re given the 𝑦-value as two. We can plug that into the equation of the line and then find 𝑥.

So let’s begin by finding the slope of segment 𝐵𝐶. So our formula to find a slope is the change in 𝑦 divided by the change in 𝑥. The change in 𝑦 is where we will take 𝑦 two and then subtract 𝑦 one and then divide by 𝑥 two minus 𝑥 one.

So to find the slope of segment 𝐵𝐶, we will let the point 𝐵 be 𝑥 one, 𝑦 one and the point 𝐶 𝑥 two, 𝑦 two. So on the numerator, we have negative five minus negative four, giving us negative one. And on the denominator, seven minus negative three, that gives us positive 10. So the slope of segment 𝐵𝐶 is negative one-tenth.

And we know that the slope of 𝐵𝐶 will be the exact same as the slope of segment 𝐴𝐵, which would be the exact same slope as segment 𝐴𝐶 because they are all colinear; they all lie on the same line, those points.

So now we need to solve for 𝑏. 𝑏- 𝑏 is the 𝑦-intercept where crosses the 𝑦-axis, where our line crosses the 𝑦-axis. And in order to find 𝑏, we need to plug an 𝑥 and a 𝑦 in. We can use point 𝐵 or point 𝐶 to plug in. Let’s use point 𝐵.

So 𝐵 is negative three, negative four. So negative three is 𝑥, and negative four is 𝑦. Let’s substitute those into our equation of the line and solve for 𝑏. So here we’ve done so.

So let’s first begin by multiplying negative one-tenth times negative three. That gives us positive three-tenths. So now, so we have positive three tenths, and let’s get rid of the denominator. Let’s multiply everything by 10. This way, on the three tenths, the 10s cancel and there’s no more denominator.

So to solve for 𝑏, let’s subtract three from both sides of the equation. The threes cancel, and now we need to divide both sides of the equation by 10. This means that 𝑏 is equal to negative forty-three tenths.

So now we have the equation of this line that all of the points 𝐴, 𝐵, and 𝐶 lie on. 𝑦 equals negative one tenth 𝑥 minus forty-three tenths. And since we know that point 𝐴 lies on this line, we can plug in two for 𝑦 and solve for the missing value 𝑥.

So let’s replace 𝑦 with two. And this will allow us to solve for 𝑥, the missing 𝑥 value of point 𝐴. So let’s begin by multiplying everything by 10, just like we did before to get rid of the denominators, resulting in 20 equals negative 𝑥 minus 43.

So now we need to add 43 to both sides of the equation, and we have 63 equals negative 𝑥. But we want 𝑥. So we need to divide both sides of the equation by negative one, resulting in 𝑥 equals negative 63, which would be our final answer.

However, let’s think about this. That’s kind of a big number. Let’s go ahead and plot 𝐵 and 𝐶 and then determine where 𝐴 should fall. Point 𝐵 is here at negative three, negative four. Point 𝐶 is here at seven, negative five. And we know that they’re collinear, so they’re all on the same line. And we know that 𝐴 is at 𝑥 comma two. So 𝑦 is two, which means 𝐴 also lies somewhere on this line.

And while this sketch is inaccurate, it does show that 𝐴 needs to be very far to the left on the 𝑥-axis. So finding that the 𝑥 value for 𝐴 is negative 63 actually seems pretty good. So once again, the value of 𝑥 is equal to negative 63.

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