### Instruction: If the straight line $$(x+y+1)+K(2x-y-1)=0$$ is perpendicular to $$2x+3y-8=0$$ then K = ?### Response: I know that two lines are perpendicular if and only if the product of their slopes is -1.
So I need to find the slopes of the given lines and set their product equal to -1.
To find the slope of a line, I can rewrite it in the form $$y=mx+b$$ where m is the slope and b is the y-intercept.
I will start with the line $$2x+3y-8=0$$ and solve for y.
Subtracting 2x from both sides, I get $$3y-8=-2x$$.
Adding 8 to both sides, I get $$3y=-2x+8$$.
Dividing by 3, I get $$y=-\frac{2}{3}x+\frac{8}{3}$$.
Now I can see that the slope of this line is $$-\frac{2}{3}$$.
I will do the same thing for the other line, $$(x+y+1)+K(2x-y-1)=0$$, but I will keep K as a variable for now.
Expanding the brackets, I get $$x+y+1+2Kx-Ky-K=0$$.
Collecting like terms, I get $$(1+2K)x+(1-K)y+(1-K)=0$$.
Subtracting $(1+2K)x$ and $(1-K)$ from both sides, I get $$(1-K)y=-(1+2K)x-(1-K)$$.
Dividing by $(1-K)$, I get $$y=-\frac{1+2K}{1-K}x-\frac{1-K}{1-K}$$.
Simplifying the fractions, I get $$y=-\frac{1+2K}{1-K}x-1$$.
Now I can see that the slope of this line is $$-\frac{1+2K}{1-K}$$.
To make the lines perpendicular, I need to have $$-\frac{2}{3} \times -\frac{1+2K}{1-K}=-1$$.
Multiplying both sides by $$-\frac{3}{2}$$, I get $$\frac{1+2K}{1-K}=\frac{3}{2}$$.
Cross-multiplying, I get $$2(1+2K)=3(1-K)$$.
Expanding, I get $$2+4K=3-3K$$.
Adding 3K to both sides, I get $$2+7K=3$$.
Subtracting 2 from both sides, I get $$7K=1$$.