TenMarks teaches you how to use the distance formula and the Pythagorean theorem.
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Learn about Distance Formula and Pythagorean Theorem In this lesson let’s learn how to use the distance formula and also the Pythagorean Theorem to find the distance between any two points in a coordinate plane. So we are given two problems, let’s do the first one first. It says, we need to find FG which is the length of this line segment and JK the length of the other line segment. And determine if they are indeed congruent which is equal. So before we start let's understand what is the distance formula. It says the distance between any two points. So if we are trying to find the distance between X and Y, right or point one and point two right, forget about X, Y. Point one and point two what it basically tells us as the way to do that is the distance between any two points on a coordinate plane right is X2 – X1² + Y2 – Y1² and the square root of this entire piece. X2 and X1 are the X coordinates of the respective two end points. And Y2 and Y1 are the Y coordinates of the two points. So let’s use this formula to find the distance FG. FG one of the coordinates, one of the coordinate of F is one and X is one, Y is two, right. The coordinate of G is X is one, two, three, four, five and Y is one, two, three, four, five right. So what is the distance? It is X2 is the second X coordinate which is the second point FG, first point and second point. Second point X coordinate is five minus first point X coordinate is one. So we square this and then we add to this Y2 – Y1 which is 5 – 2² and we take the square root of all of this. So the length FG would be 5 – 1 = 4, 4² is 16 plus 5 – 2 = 3, 3² is 9 so we’re in a 16 + 9 = 25 or root of 25 is five. So FG = 5. Let's compute JK similarly JK equals the square root of what's its coordinates? The coordinates here are for J is one, two, three, four so -4 and zero right. And coordinates for K are -1 and one, two, three so -3 right -1 or -3. So let’s substitute the values X2 is -1 minus Y2 is -4 right squared plus -1 - -4, this is X2, this is X1. Y2 is -3 minus Y1 is 0², what is this give us? -1, negative of negative is a positive plus four, right. -1 + 4 = 3. So 3² is 9, plus, -3 – 0 = -3 x ² is 9 so this is rule of 18, right. Square root of 18 is 3 times square of two. Right, because 3 x 3 = 9, 9 x 2 = 18. So the length of the line segment JK is 3 x root of two. Length of FG is five. What does that mean? That FG is not congruent to JK, right they're not equal. Remember the key thing that we learned is the formula that like find the distance between any two points is you take the X coordinates subtract them take the Y coordinate subtract them, square them individually, add them and be under it, right, square root of all of this. Let's use the same concept for the second problem. It says use the distance formula and the Pythagorean Theorem to find the distance from D to E, we are given the coordinates. So first we got to use the distance formula. What’s the distance formula just taught us? X2 – X1 right. So -2 minus X1 is 3² plus Y2 which is -5 minus Y1 is 4², right. Which is what? Square root of -2 + -3 = -5 x 5 = 25 plus -5 + -4 = -9² = 81 which is square root of 106, square root of 106 is approximately 10.3 we have to find it to the nearest 10th right. So the distance between D and E is 10.3 that’s how we use the distance formula we’re doing. All I did is take the X coordinates, right. X2, X1 subtract them so our X2 and X1 subtract them squared. Take the Y2 and Y1, subtract them and square we get 25 + 81 = 106 and we take the square root of this entire thing to give us 10.3. Now, there's another way to do this, if we look at this line, we can form it’s a right angle, rectangle. This is the right angle right. The Pythagorus Theorem basically says that if we have a right triangle and then the length of one side square which is A² + B² right, equals the length of this line which is C², this is called the hypothenuse, which is the longest line in the right angle-triangle, right. So if this is the