Calculate: 3.2.1 The coordinates of the midpoint of AC...

Calculate: 3.2.1 The coordinates of the midpoint of AC (2) 3.2.2 The gradient of line AC (2) 3.2.3 The magnitude of ABX (2) 3.2.4 The distance AB (2) 3.2.5 The equation of a line through A, perpendicular to BC (4) [24]

Transcript

00:01 So for this question, first of all, it says that find out the coordinate of the midpoint of ac.

00:06 So the midpoint of two points will be x1 plus x2 divided by 2 and y1 plus y2 divided by 2.

00:18 So ac, a is at 2 3 and c is at 5 negative 3.

00:32 So the midpoint can be calculated by 2 plus 5 divided by 2 and 3 plus negative 3 divided by 2.

00:46 So this will give you 3 .5 and 0.

00:51 So this will be the midpoint of ac.

00:54 2.

00:55 The gradient of line ac.

01:00 Gradient formula basically can be calculated by a slope.

01:06 So m equals y2 minus y1 divided by x2, divided by x2, divide by x1.

01:13 So again, we're talking about ac, so the gradient equals negative 3 minus 3, divide by 5 minus 2, negative 2, 3.

01:37 The magnitude of ab, the matter of abx.

01:46 X basically means that the vector ab.

01:51 So vector ab basically is b minus a.

02:01 So basically b is negative 1, 0.

02:07 So negative 1 minus 2, which is a and 0 minus 3.

02:14 So ab is negative 3, negative 3, vector.

02:19 Then same idea, let's calculate ac, which is c minus a.

02:26 C is 5 minus 2 and native 3 minus 3.

02:35 So the vector ac is 3 -native 6.

02:38 All right, with that in mind, let's say ab vector cross ac is abx, multiply by acx minus aby, multiply by acy, multiply by acy, an absolute value.

03:16 So let's plug in all the value.

03:18 A, bx, negative 3, acx, and a, b, y is negative 3, multiply by acy negative 6...

Question

Calculate: 3.2.1 The coordinates of the midpoint of AC (2) 3.2.2 The gradient of line AC (2) 3.2.3 The magnitude of ABX (2) 3.2.4 The distance AB (2) 3.2.5 The equation of a line through A, perpendicular to BC (4) [24]

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