• 注意，这是对OP的original question的回答;此后，他们对自己的问题进行了大幅修改。

这是@dcarlson的解决方案here的扩展，添加lm()来从点的向量中找到截距和斜率。

intersection <- function(x, y1, y2) {
  l1 <- unname(coef(lm(y1 ~ x)))
  l2 <- unname(coef(lm(y2 ~ x)))
  x_int <- (l2[1] - l1[1]) / (l1[2] - l2[2])
  y_int <- l1[1] + l1[2] * x_int
  c(x = x_int, y = y_int)
}

# intersection of m2 and m5
m2_m5 <- intersection(l, m[2, ], m[5, ])
m2_m5
#        x         y 
# 0.720027 56.960115 

# intersection of m3 and m5
m3_m5 <- intersection(l, m[3, ], m[5, ])
m3_m5
#         x         y 
# -25.59092 -21.97391

m3和m5的交点不在你的绘图区域内，但我们可以绘制m2和m5的交点：

plot(l, m[2, ], type = "l", lwd = 2, col = 2, ylim = c(50, 90))
lines(l, m[3, ], type = "l", lwd = 2, col = 3)
lines(l, m[5, ], type = "l", lwd = 2, col = 5)
legend("topleft", legend = c("m2", "m3", "m5"), col = c(2, 3, 5), lty = 1, lwd = 2)
points(m2_m5[[1]], m2_m5[[2]], cex = 2, col = 2, lwd = 2)