![Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, - 2) . Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, - 2) .](https://haygot.s3.amazonaws.com/questions/1175717_1348145_ans_49de745563ab42b7a74039c45baa4430.jpg)
Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, - 2) .
Kumon Hi-Strip 5 Bacolod Center - Fun Trivia! Do you know what shape the implicit curve equation (x2+y2-1)3-x2y3=0 produces? | Facebook
Solve: (x + y)^2/3 + 2(x - y)^2/3 = 3(x^2 - y^2)^1/3 and 3x -2y = 13. - Sarthaks eConnect | Largest Online Education Community
![problem solving - Decipher the greeting card $( X^2 +Y^2 -1 ) ^3 - X^2 Y^3 = 0$ - Mathematics Stack Exchange problem solving - Decipher the greeting card $( X^2 +Y^2 -1 ) ^3 - X^2 Y^3 = 0$ - Mathematics Stack Exchange](https://i.stack.imgur.com/eXoBX.png)
problem solving - Decipher the greeting card $( X^2 +Y^2 -1 ) ^3 - X^2 Y^3 = 0$ - Mathematics Stack Exchange
![UberFacts on Twitter: "You can plot a heart on a graph using the equation: ( x2 + y2 − 1)3 − x2y3 = 0 https://t.co/ioHJXHXgVp" / Twitter UberFacts on Twitter: "You can plot a heart on a graph using the equation: ( x2 + y2 − 1)3 − x2y3 = 0 https://t.co/ioHJXHXgVp" / Twitter](https://pbs.twimg.com/media/DtMPrKNWwAEQzRz.jpg)