Lines 𝐴𝐶 and 𝐵𝐷 are the diagonals of a square. The equation of line 𝐵𝐷 is 𝑦 minus three 𝑥 equals four. Find the equation of line 𝐴𝐶.So firstly, what we’re gonna do is actually find the centre of our square. And I’ve labelled the centre of our square 𝐸 in our diagram. We also know that there’s actually a right angle at 𝐸. And we have a right angle it’s because we know that 𝐴𝐶 and 𝐵𝐷 are the diagonalsof a square. And so, therefore, 𝐴𝐶 and 𝐵𝐷 are perpendicular. And this is a relationship that’s gonna be very useful later on. So first of all, let’s find the coordinates of point 𝐸 because this is the pointthat’s gonna be on both of our lines because it’s the centre of our square. So straight away, we know that the 𝑥-coordinate of point 𝐸 is gonna be zero. So 𝑥 is equal to zero. And that’s because it’s on the 𝑦-axis. So then what we can do is actually substitute this into 𝑦 minus three 𝑥 equals fourto actually give us the value of 𝑦 at this point. So therefore, if we do, we get 𝑦 minus three multiplied by zero equals four. So therefore, we’re gonna get 𝑦 is equal to four because 𝑦 minus three multipliedby zero, well three multiplied by zero is just zero. So 𝑦 minus zero equals four. So therefore, 𝑦 must be equal to four. So therefore, we know that the coordinates of the centre of our square, so point 𝐸,are zero, four. Okay, great! So what do we do now? So now we actually move on to line 𝐴𝐶 because what the question wants us to do isfind the equation of line 𝐴𝐶. So first of all, to actually find the equation of line 𝐴𝐶, we’re gonna have a lookat line 𝐵𝐷. So we’ve got line 𝐵𝐷 is 𝑦 minus three 𝑥 is equal to four. Well by actually adding three 𝑥 to each side, what we can actually do is rearrangeit into 𝑦 equals three 𝑥 plus four. But why would we want to do that? Well, we do that because, actually, it gives it in the form 𝑦 equals 𝑚𝑥 plus𝑐. And this is the general form for the equation of a straight line, where 𝑚 is ourgradient and 𝑐 is our 𝑦 intercept. Okay, great! So this is gonna be be really useful. First of all actually, we’ll just go back to something we’ve already done, because ifwe know that 𝑐 is our 𝑦-intercept, then therefore it tells us that positive fouris gonna be our 𝑦-intercept. And actually yet if we look back at our point 𝐸, we can actually see that the𝑦-intercept, so the 𝑦-value where it crosses the 𝑦-axis, is actually four. So yes that was correct. So now it’s actually the gradient that’s gonna help us find the equation of line𝐴𝐶. And that’s because of the relationship we actually highlighted earlier, because wesaid that 𝐴𝐶 and 𝐵𝐷 are perpendicular to one another. So they’re right angles. And we actually know that if two lines are perpendicular to each other, theirgradients, which I’ve called here 𝑚 one and 𝑚 two, when they’re multipliedtogether are equal to negative one. And another way we can actually say that is that actually the gradient of a line isthe negative reciprocal of the gradient of the line that is perpendicular to it. Okay, great! So let’s use this to find the equation of line 𝐴𝐶.So we can now say that line 𝐴𝐶, if we’re gonna have it in the form 𝑦 equals 𝑚𝑥plus 𝑐, our 𝑚, our gradient, is going to be negative a third. And the reason we know it’s negative a third is cause if we look back at the line𝐵𝐷, we can see that the gradient of the line 𝐵𝐷 was three. And therefore, negative a third multiplied by three is gonna give us negative threeover three, which gives us negative one. And again, using the other definition, we can see that, well, negative one over threeor negative a third is actually that negative reciprocal of three. Okay, great! So we’ve now found the gradient. What about the 𝑦-intercept? Well, from earlier, we’ve already found out the 𝑦-intercept is equal to four. And that’s because four is the value of 𝑦 where the centre of the square is, causethat’s where the diagonals actually meet and that’s on both lines. So then we can actually prove that by actually substituting our point 𝐸 values intoour equation because we’ve got 𝑦 is equal to negative a third 𝑥 plus 𝑐. And we get negative a third 𝑥 because 𝑚 is equal to negative a third. So then if we substitute 𝑦 is equal to four and 𝑥 is equal to zero into that, weget four is equal to negative third multiplied by zero plus 𝑐. So therefore, we get four is equal to 𝑐. So as we’ve said, 𝑐 is equal to four cause that’s our 𝑦-intercept. So therefore, we can say that the equation of line 𝐴𝐶 is 𝑦 equals negative a third𝑥 plus four.