## to find co-cordinate of Intersection point of line?

### S (view profile)

on 4 Apr 2013
```x=[7.8 8.25 8.5];
y=[0.96 0.99 0.94];
p=polyfit(x,y,2);
f=polyval(p,x);
plot(x,y,'o',x,f,'-');
hold on;
line([7.8 8.5],[0.96 0.94]);
hold on;
line([8.25 8.25],[0 0.99]);
```

Now want to compute intersection point of both lines?

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### Walter Roberson (view profile)

Answer by Walter Roberson

### Walter Roberson (view profile)

on 4 Apr 2013

```p1 = polyfit(x(1:2), y(1:2), 2);
b1 = polyval(p1, 0);
m1 = polyval(p1, 1) - b1;
```
```p2 = polyfit(x(2:3), y(2:3), 2);
b2 = polyval(p2, 0);
m2 = polyval(p1, 1) - b2;
```

And now you have m1*x + b1 = m2*x + b2, so m1 * x - m2*x = b2 - b1, so (m1 - m2) * x = b2 - b1, and so x = (b2 - b1) / (m1 - m2), after which y = m1*x + b1 to find the corresponding y.

S

### S (view profile)

on 6 Apr 2013
```x1=7.8;
x2=8.5;
y1=0.96;
y2=0.94;
p1 = polyfit([x1 x2], [y1 y2], 2);
b1= polyval(p1,1);
m1=polyval(p1,2)-b1;
x3=8.25;
x4=8.25;
y3=0;
y4=.99;
p2 = polyfit([x3 x4], [y3 y4], 2);
b2 = polyval(p2, 1);
m2 = polyval(p2, 2) - b2;
```

I got x value = -1.2867; from which co-ordinate this value corresponds to? Actually I want to compute intersection of two line with respect to x=[7.8 8.25 8.5]; y=[0.96 0.99 0.94]; over which those two lines are plotted?

```x=[7.8 8.25 8.5];
y=[0.96 0.99 0.94];
p=polyfit(x,y,2);
f=polyval(p,x);
plot(x,y,'o',x,f,'-');
hold on;
line([7.8 8.5],[0.96 0.94]);
hold on;
line([8.25 8.25],[0 0.99]);
```

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