Can someone check my work please?
Convert each polar equation to a rectangular equation. Then determine the graph's slope and y-intercept.
( I'm going to type 'x' instead of theta )
r sin(x- pi/4)=2
r(sin(x- pi/4))=2
r(sinx*cos pi/4 - sin pi/4*cosx)=2
r(sqrt2/2*sin x - sqrt2/2*cos x)=2
sqrt2/2*rsin x - sqrt2/2*rcosx=2
(sqrt2/2*y - sqrt2/2*x=2)^2
1/2(y^2) - 1/2(x^2)= 4
1/2(y^2 - x^2)=4
y^2 - x^2= 8
(y^2= 8 + x^2) sqrt
y= sqrt(8) + x
slope= 1
y-int= sqrt 8
miraculously, you made two major errors that cancelled each other out, and you came up with the right answer.
line 4 can be reduced to √2y - √2x = 4
which convert to y = -x + √8
so your answer is right
why are you squaring the equation in line 5???, you had it!!!
When you square an equation you CANNOT just square each term.
Your left side should have contained 3 terms.
Then in line 9 you cannot take the square root of each term!
eg. √(16+9) is not equal to √16 + √9
To convert the polar equation to a rectangular equation, you correctly applied the conversions and simplifications up to line 4. However, in line 5, you made a mistake by squaring the equation. When you square an equation, you need to square all terms, not just some of them.
The correct equation in line 5 should be:
(sqrt2/2 * y - sqrt2/2 * x)^2 = 4
Expanding this equation, we get:
(2/4) * (2y^2 - 2xy + 2x^2) = 4
Simplifying further, we have:
y^2 - xy + x^2 = 2
To determine the slope and y-intercept of the graph, we need to rearrange the equation in the standard form of a line, y = mx + b (where m is the slope and b is the y-intercept).
Rearranging the equation, we get:
y^2 - xy + x^2 - 2 = 0
To isolate y, consider treating this equation as a quadratic equation in y. Applying the quadratic formula, we have:
y = (x ± √(4 - 4(x^2 - x^2 + 8))) / 2
y = (x ± √(32)) / 2
y = (x ± 4√2) / 2
y = 2x ± 2√2
From this, we can see that the slope (m) is 2, and the y-intercept (b) can take two values, +2√2 and -2√2.
Therefore, the correct slope is 2 and the y-intercept can be either +2√2 or -2√2.